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Full text of "Math 380 Elementary Algebra Textbook"

Math 380 Elementary Algebra Textbook 

Spring 2012 Edition 

Department of Mathematics 
College of the Redwoods 

January 13, 2012 



Copyright 

All parts of this prealgebra textbook are copyrighted © 2011 in the 
name of the Department of Mathematics, College of the Redwoods. They 
are not in the public domain. However, they are being made available 
free for use in educational institutions. This offer does not extend to any 
application that is made for profit. Users who have such applications 
in mind should contact David Arnold at david-arnold@redwoods.edu or 
Bruce Wagner at bruce-wagner@redwoods.edu. 

This work is licensed under the Creative Commons Attribution-Non- 
Commercial-NoDerivs 3.0 Unported License. To view a copy of this li- 
cense, visit http://creativecommons.Org/licenses/by-nc-nd/3.0/ or send 
a letter to Creative Commons, 171 Second Street, Suite 300, San Fran- 
cisco, California, 94105, USA. 



Contents 



The Arithmetic of Numbers 1 

1.1 An Introduction to the Integers 2 

The Integers 3 

Absolute Value 4 

Integer Addition 5 

Mathematical Properties of Addition 6 

Integer Subtraction 7 

Integer Multiplication 8 

Mathematical Properties of Multiplication 9 

Exponents 10 

Graphing Calculator: Negating versus Subtracting 11 

Exercises 14 

Answers 15 

1.2 Order of Operations 16 

Grouping Symbols 19 

Absolute Value Bars as Grouping Symbols 20 

Nested Grouping Symbols 20 

Evaluating Algebraic Expressions 21 

Evaluating Fractions 22 

Using the Graphing Calculator 23 

Exercises 27 

Answers 30 

1.3 The Rational Numbers 31 

Reducing Fractions to Lowest Terms 31 

Multiplying Fractions 33 

Dividing Fractions 35 

Adding Fractions 36 

Order of Operations 37 

Fractions on the Graphing Calculator 39 

Exercises 43 

Answers 46 



CONTENTS 



1.4 Decimal Notation 48 

Adding and Subtracting Decimals 49 

Multiplication and Division of Decimals 51 

Order of Operations 52 

Rounding Using the Graphing Calculator 55 

Exercises 57 

Answers 59 

1.5 Algebraic Expressions 60 

The Distributive Property 61 

Speeding Things Up a Bit 62 

Distributing a Negative Sign 63 

Combining Like Terms 64 

Order of Operations 66 

Exercises 69 

Answers 70 

Solving Linear Equations and Inequalities 73 

2.1 Solving Equations: One Step 74 

Equivalent Equations 74 

Wrap and Unwrap, Do and Undo 75 

Operations that Produce Equivalent Equations 76 

More Operations That Produce Equivalent Equations .... 79 

Writing Mathematics 82 

Exercises 84 

Answers 85 

2.2 Solving Equations: Multiple Steps 87 

Variables on Both Sides of the Equation 90 

Simplifying Expressions When Solving Equations 92 

Exercises 96 

Answers 97 

2.3 Clearing Fractions and Decimals 99 

Canceling is More Efficient 100 

Clearing Fractions from an Equation 101 

Clearing Decimals from an Equation 106 

Exercises 109 

Answers 110 

2.4 Formulae Ill 

Clearing Fractions 113 

Geometric Formulae 115 

Exercises 118 

Answers 120 

2.5 Applications 121 

Exercises 132 

Answers 134 

2.6 Inequalities 135 

Ordering the Real Numbers 136 



CONTENTS 



Set-Builder Notation 136 

Interval Notation 138 

Equivalent Inequalities 139 

Reversing the Inequality Sign 141 

Multiple Steps 143 

Summary Table of Set-Builder and Interval Notation 146 

Exercises 147 

Answers 149 

Introduction to Graphing 151 

3.1 Graphing Equations by Hand 152 

The Cartesian Coordinate System 152 

Plotting Ordered Pairs 154 

Equations in Two Variables 156 

Graphing Equations in Two Variables 157 

Guidelines and Requirements 159 

Using the TABLE Feature of the Graphing Calculator .... 160 

Exercises 165 

Answers 168 

3.2 The Graphing Calculator 171 

Reproducing Calculator Results on Homework Paper .... 174 

Adjusting the Viewing Window 176 

Exercises 179 

Answers 180 

3.3 Rates and Slope 183 

Measuring the Change in a Variable 184 

Slope as Rate 185 

The Steepness of a Line 189 

The Geometry of the Slope of a Line 192 

Exercises 196 

Answers 199 

3.4 Slope-Intercept Form of a Line 201 

Applications 206 

Exercises 209 

Answers 212 

3.5 Point-Slope Form of a Line 213 

Parallel Lines 217 

Perpendicular Lines 219 

Applications 222 

Exercises 224 

Answers 225 

3.6 Standard Form of a Line 228 

Slope- Intercept to Standard Form 230 

Point-Slope to Standard Form 232 

Intercepts 234 

Horizontal and Vertical Lines 237 



CONTENTS 

Exercises 240 

Answers 242 

Systems of Linear Equations 245 

4.1 Solving Systems by Graphing 246 

Exceptional Cases 250 

Solving Systems with the Graphing Calculator 253 

Exercises 259 

Answers 260 

4.2 Solving Systems by Substitution 262 

Exceptional Cases Revisited 268 

Exercises 270 

Answers 272 

4.3 Solving Systems by Elimination 273 

Exceptional Cases 276 

Exercises 279 

Answers 280 

4.4 Applications of Linear Systems 282 

Exercises 290 

Answers 291 

Polynomial Functions 293 

5.1 Functions 294 

Mapping Diagrams 296 

Function Definition 297 

Mapping Diagram Notation 299 

Function Notation 301 

Interchanging y and f{x) 303 

Exercises 305 

Answers 307 

5.2 Polynomials 308 

Ascending and Descending Powers 309 

The Degree of a Polynomial 311 

Polynomial Functions 312 

The Graph of a Polynomial Function 313 

Exercises 317 

Answers 319 

5.3 Applications of Polynomials 321 

Zeros and ^-intercepts of a Function 325 

Exercises 330 

Answers 333 

5.4 Adding and Subtracting Polynomials 334 

Negating a Polynomial 335 

Subtracting Polynomials 336 

Some Applications 338 

Exercises 340 



CONTENTS vii 

Answers 342 

5.5 Laws of Exponents 343 

Multiplying With Like Bases 344 

Dividing With Like Bases 345 

Raising a Power to a Power 347 

Raising a Product to a Power 349 

Raising a Quotient to a Power 350 

Exercises 353 

Answers 355 

5.6 Multiplying Polynomials 356 

The Product of Monomials 356 

Multiplying a Monomial and a Polynomial 357 

Multiplying Polynomials 359 

Speeding Things Up a Bit 360 

Some Applications 362 

Exercises 366 

Answers 368 

5.7 Special Products 370 

The FOIL Method 370 

The Difference of Squares 373 

Squaring a Binomial 375 

An Application 378 

Exercises 380 

Answers 382 

6 Factoring 385 

6.1 The Greatest Common Factor 386 

Finding the Greatest Common Factor of Monomials 388 

Factor Out the GCF 390 

Speeding Things Up a Bit 392 

Factoring by Grouping 394 

Exercises 396 

Answers 398 

6.2 Solving Nonlinear Equations 399 

Linear versus Nonlinear 400 

Using the Graphing Calculator 404 

Exercises 410 

Answers 411 

6.3 Factoring ax 2 + bx + c when a = 1 413 

The ac-Method 413 

Speeding Things Up a Bit 416 

Nonlinear Equations Revisited 417 

Exercises 424 

Answers 425 

6.4 Factoring ax 2 + bx + c when a^ 1 427 

Speeding Things Up a Bit 428 



i CONTENTS 

Nonlinear Equations Revisited 431 

Exercises 437 

Answers 438 

6.5 Factoring Special Forms 440 

Perfect Square Trinomials 440 

The Difference of Squares 444 

Factoring Completely 446 

Nonlinear Equations Revisited 447 

Exercises 452 

Answers 454 

6.6 Factoring Strategy 456 

Using the Calculator to Assist the ac-Method 462 

Exercises 464 

Answers 465 

6.7 Applications of Factoring 467 

Exercises 474 

Answers 475 

Rational Expressions 477 

7.1 Negative Exponents 478 

Laws of Exponents 480 

Raising to a Negative Integer 482 

Applying the Laws of Exponents 484 

Clearing Negative Exponents 485 

Exercises 488 

Answers 490 

7.2 Scientific Notation 492 

Multiplying Decimal Numbers by Powers of Ten 493 

Scientific Notation Form 495 

Placing a Number in Scientific Notation 495 

Scientific Notation and the Graphing Calculator 497 

Exercises 502 

Answers 504 

7.3 Simplifying Rational Expressions 505 

Multiplying and Dividing Rational Expressions 505 

Adding and Subtracting Rational Expressions 507 

The Least Common Denominator 508 

Dividing a Polynomial by a Monomial 511 

Exercises 513 

Answers 514 

7.4 Solving Rational Equations 516 

Solving Rational Equations with the Graphing Calculator . . 518 

Numerical Applications 521 

Exercises 523 

Answers 524 

7.5 Direct and Inverse Variation 525 



CONTENTS 



Inversely Proportional 527 

Exercises 531 

Answers 533 

8 Quadratic Functions 535 

8.1 Introduction to Radical Notation 536 

Using the Graphing Calculator 539 

Approximating Square Roots 542 

Exercises 545 

Answers 546 

8.2 Simplifying Radical Expressions 547 

Simple Radical Form 548 

The Pythagorean Theorem 549 

Proof of the Pythagorean Theorem 550 

Applications 553 

Exercises 555 

Answers 557 

8.3 Completing the Square 558 

Perfect Square Trinomials Revisited 561 

Completing the Square 562 

Solving Equations by Completing the Square 563 

Exercises 568 

Answers 569 

8.4 The Quadratic Formula 571 

Exercises 580 

Answers 582 

Index 585 



Chapter 1 

The Arithmetic of Numbers 



In 1960, Belgian Geologist Jean de Heinzelein de Braucourt discovered the Is- 
liango bone in central Africa. This bone, dated to be more than 20,000 years 
old, is believed to be the oldest known artifact indicating the use of arith- 
metic. The first written records indicate the Egyptians and Babylonians used 
arithmetic as early as 2000 BC. The Mayans used arithmetic to make astro- 
nomical computations, and developed the concept of zero over 2,000 years ago. 
The word "arithmetic" is derived from the Greek word arithmos (translated as 
"number" ) . It is the oldest and most elementary branch of mathematics and 
is used for a variety of tasks ranging from simple counting to advanced science 
and business calculations. 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



1.1 An Introduction to the Integers 

We begin with the set of counting numbers, formally called the set of natural 
numbers. 




If we add the number zero to the set of natural numbers, then we have a set 
of numbers that are called the whole numbers. 



The Whole Numbers. The set 




W={0,1,2,3,4,5,. 


••} 


is called the set of whole numbers. 





The number is special, in that whenever you add it to another whole number, 
you get the identical number as an answer. 

Additive Identity Property. If a is any whole number, then 

a + = a. 
For this reason, the whole number is called the additive identity. 

Thus, for example, 3 + = 3, 15 + = 15, and 123 + = 123. These are all 
examples of the additive identity property. 

Every natural number has an opposite, so that when you add them together, 
their sum is zero. 

Additive Inverse Property. If a is any natural number, then define the 
opposite of a, symbolized by —a, so that 

a + (-a) = 0. 

The number —a is called the "opposite of a," or more formally, the additive 
inverse of a. 

For example, the opposite (additive inverse) of 3 is —3, and 3 + (—3) = 0. The 
opposite (additive inverse) of 12 is —12, and 12 + (—12) = 0. The opposite of 



1.1. AN INTROD UCTION TO THE INTEGERS 



254 is —254, and 254+ (—254) = 0. These are all examples of additive inverses 
and the additive inverse property. 

Because 7 + (—7) = 0, we've said that —7 is the opposite (additive inverse) 
of 7. However, we can also turn that around and say that 7 is the opposite 
of —7. If we translate the phrase "the opposite of —7 is 7" into mathematical 
symbols, we get —(—7) = 7. 




Thus, for example, -(-11) = 11, -(-103) = 103, and -(-1255) = 1255. 

The Integers 

If we collect all the natural numbers and their additive inverses, then include 
the number zero, we have a collection of numbers called the integers. 



The Integers. The set 












Z = {...,-5, 


-4, 


-3, 


-2, 


-1,0,1,2,3,4,5,. 


■■} 


is called the set of integers. 













The integers can be made to correspond to points on a line in a very natural 
manner. First, draw a line, then locate the number zero anywhere you wish. 
Secondly, place the number one to the right of zero. This determines the length 
of one unit. Finally, locate the numbers 1, 2, 3, 4, 5, . . . to the right of zero, 
then their opposites (additive inverses) —1, —2, —3, —4, —5, . . . to the left of 
zero (see Figure 1.1). 

<H 1 1 1 1 1 1 1 1 1 h- ► 

-5-4-3-2-10 1 2 3 4 5 

Figure 1.1: Each integer corresponds to a unique position on the number line. 

Note that as we move to the right on the number line, the integers get larger. 
On the other hand, as we move to the left on the number line, the integers get 
smaller. 

Positive and negative integers. On the number line, some integers lie to 
the right of zero and some lie to the left of zero. 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



• 


If a is an 
integer. 


integer 


that lies to the right of zero 


, then a 


is called 


a positive 


• 


If a is an 
integer. 


integer 


that lies to the left of zero, 


then a 


is called s 


negative 



Thus, 4, 25, and 142 are positive integers, while —7, —53, and —435 are negative 
integers. 

Absolute Value 

The absolute value (or magnitude) of an integer is defined as follows. 

The Absolute Value of an Integer. If a is an integer, then the absolute 
value of a, written \a\, is defined as the distance between the integer and zero 
on the number line. 



You Try It! 



Simplify: | - 23 1 



EXAMPLE 1. Simplify | - 4|. 

Solution: Consider the position of —4 on the number line. Note that —4 lies 
four units away from zero. 



4 units 



-«-H 



H h 



H 1 1 h 



H h 



+->■ 



-1 1 



Answer: I - 23| = 23 



Because the absolute value (magnitude) of an integer equals its distance from 
zero, | - 4| = 4. 

□ 

In similar fashion: 

• The integer 5 lies five units away from zero. Hence, |5| = 5. 

• The integer lies zero units away from zero, Hence, |0| = 0. 

Note that the absolute value of any number is either positive or zero. That is, 
the absolute value of a number is nonnegative (not negative). 



1.1. AN INTROD UCTION TO THE INTEGERS 



Integer Addition 

This section is designed to provide a quick review of integer addition. For a 
more thorough introduction to integer addition, read section two of chapter 
two of our prealgebra textbook, provided online at the following URL: 

http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/ 

chapter2 .pdf 

We consider the first of two cases. 



Adding Integers with Like Signs. To add two integers with like signs (both 
positive or both negative), add their magnitudes (absolute values), then prefix 
their common sign. 



You Try It! 



EXAMPLE 2. Simplify 7+12. 

Solution: We have like signs. The magnitudes (absolute values) of 7 and 12 
are 7 and 12, respectively. If we add the magnitudes, we get 19. If we prefix 
the common sign, we get 19. That is: 

7 + 12= 19 



Simplify: 13 + 28 



Answer: 41 



□ 



You Try It! 



EXAMPLE 3. Simplify -8+ (-9). 

Solution: We have like signs. The magnitudes (absolute values) of —8 and 
—9 are 8 and 9, respectively. If we add the magnitudes, we get 17. If we prefix 
the common sign, we get —17. That is: 



Simplify: —12 



-21) 



+ (-9) 



17 



Answer: 



-33 



□ 



Next, we consider the case where we have unlike signs. 



Adding Integers with Unlike Signs. To add two integers with unlike 
signs (one positive and one negative), subtract the integer with the smaller 
magnitude (absolute value) from the number with the larger magnitude, then 
prefix the sign of the integer with the larger magnitude. 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: 12 



-29) 



EXAMPLE 4. Simplify -14+11. 

Solution: We have unlike signs. The magnitudes (absolute values) of —14 and 
11 are 14 and 11, respectively. If we subtract the smaller magnitude from the 
larger, we get 3. The number —14 has the larger magnitude, so we prefix our 
answer with its negative sign. That is: 



Answer: —17 



-14+11 



□ 



You Try It! 



Simplify: 32 + (-90) 



EXAMPLE 5. Simplify 40 + (-25). 

Solution: We have unlike signs. The magnitudes (absolute values) of 40 and 
—25 are 40 and 25, respectively. If we subtract the smaller magnitude from the 
larger, we get 15. The number 40 has the larger magnitude, so we prefix our 
answer with its positive sign. That is: 



Answer: 



-58 



40 +(-25) = 15 



□ 



Mathematical Properties of Addition 

The order in which we add integers does not matter. That is, —20 + 34 gives 
an answer identical to the sum 34 + (—20). In both cases, the answer is 14. 
This fact is called the commutative property of addition. 




Next, when we add three integers, it does not matter which two we add 
first. For example, if we add the second and third of three numbers first, we 
get: 



-11 + (-2 + 5) = -11 + 3 



Parentheses first: 
Add: -11 + 3 = 



1.1. AN INTRODUCTION TO THE INTEGERS 7 

On the other hand, if we add the first and second of three numbers first, we 
get: 

(-11 + (-2)) + 5 = -13 + 5 Parentheses first: -11 + (-2) = -13 
= -8 Add: -13 + 5 = -8 

Thus, —11 + (—2 + 5) = (—11 + (—2)) + 5. This fact is called the associative 
property of addition. 



The Associative Property of Addition. If a, 6, and c are any three integers, 
then: 

a + (b + c) = {a + b) + c 



Integer Subtraction 

Subtraction is the inverse, or the opposite, of addition. 




You Try It! 



EXAMPLE 6. Simplify: -13-27 Simplify: -11-15 

Solution: The "opposite" (additive inverse) of 27 is —27. So, subtracting 27 
is the same as adding —27. 

-13 - 27 = -13 + (-27) Subtracting 27 is the same 

as adding —27. 

= —50 Add the magnitudes, then 

prefix the common negative sign. 

Answer: —26 



You Try It! 



□ 



EXAMPLE 7. Simplify: -27- (-50) Simplify: -18 - (-54) 



Answer: 36 



You Try It! 



Simplify: (-18)(-5) 



Answer: 90 



8 CHAPTER 1. THE ARITHMETIC OF NUMBERS 

Solution: The "opposite" (additive inverse) of —50 is —(—50), or 50. So, 
subtracting —50 is the same as adding 50. 



-27- (-50) = -27 + 50 



23 



Subtracting —50 is the same 
as adding 50. 

Subtract the smaller magnitude from 
the larger magnitude, then prefix the 
sign of the larger magnitude. 



□ 



Integer Multiplication 

This section is designed to provide a quick review of multiplication and division 
of integers. For a more thorough introduction to integer multiplication and 
division, read section four of chapter two of our prealgebra textbook, provided 
online at the following URL: 

http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/ 

chapter2 .pdf 



Like Signs. If a 


and 6 


are integers with like 


signs (both positive 


or 


both 


negative), then the product ab and the quotient 


a/b 


are positive. 






(+)(+) = 


= + 


or 




(+)/(+) = + 






(-)(-) = 


= + 


or 




(")/(-) = + 







EXAMPLE 8. Simplify each of the following expressions: 
(a) (2)(3) (b) (-12)(-8) (c) -14/(-2) 

Solution: When multiplying or dividing, like signs yield a positive result. 
(a)(2)(3)=6 (b) (-12)(-8) = 96 (c) -14/(-2) = 7 



□ 



Unlike Signs. If a and b are integers with unlike signs (one positive 


and 


one 


negative), then the product 


ab and the quotient 


a/b are negative. 






(+)(") = - 


or 


(+)/(") = - 






(")(+) = - 


or 


(")/(+) = - 







1.1. AN INTROD UCTION TO THE INTEGERS 



You Try It! 



EXAMPLE 9. Simplify each of the following expressions: 
(a) (2)(-12) (b) (-9)(12) (c) 24/(-8) 

Solution: When multiplying or dividing, unlike signs yield a negative result, 
(a) (2) (-12) = -24 (b) (-9) (12) = -108 (c) 24/(-8) = -3 



Simplify: (-19)(3) 



Answer: —57 



Mathematical Properties of Multiplication 

The order in which we multiply integers does not matter. That is, (— 8)(5) 
gives an answer identical to (5) (—8). In both cases, the answer is —40. This 
fact is called the commutative property of multiplication. 



The Commutative Property of Multiplication. If a and b are any two 


integers, then: 


a ■ b = b ■ a 



Next, when we multiply three integers, it does not matter which two we 
multiply first. If we multiply the second and third of three numbers first, we 
get: 

(-3) [ (-4)(-5) ] = (-3) (20) Brackets first: (-4)(-5) = 20 

= -60 Multiply: (-3)(20) = -60 

On the other hand, if we multiply the first and second of three numbers first, 
we get: 

[ (-3)(-4) ] (-5) = (12) (-5) Brackets first: (-3)(-4) = 12 

= -60 Multiply: (12)(-5) = -60 

Thus, (— 3) [(— 4)(— 5)] = [(— 3)(— 4)] (— 5). This fact is called the associative 
property of multiplication. 

The Associative Property of Multiplication. If o, 6, and c are any three 
integers, then: 

a ■ (b ■ c) = (a ■ b) ■ c 



□ 



When you multiply an integer by 1, you get the identical number back as 
the product. For example, (1)(5) = 5 and ( — 11) (1) = —11. This fact is known 
as the multiplicative identity property. 



10 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



The Multiplicative Identity Property. If a is any integer, then: 

1 ■ a = a and a ■ 1 = a 
For this reason, the integer 1 is called the "multiplicative identity." 

Finally, note that (— 1)(5) = —5. Thus, multiplying 5 by —1 is identical to 
taking the "opposite" of 5 or negating 5. 

The Multiplicative Property of — 1. Multiplying by minus one is identical 
to negating. That is: 

(— l)a = —a 



Exponents 

In the exponential expression a™, the number a is called the base, while the 
number n is called the exponent. We now define what is meant by an exponent. 

Exponents. Let a be an integer and let n be any whole number. If n ^ 0, 
then: 



a = a ■ a ■ a ■ 



a 



n times 

That is, to calculate a™, write a as a factor n times. 



You Try It! 



Simplify: (-2) 



EXAMPLE 10. Simplify (-2) 3 . 

Solution: In the exponential expression (— 2) 3 , note that —2 is the base, while 
3 is the exponent. The exponent tells us to write the base as a factor three 
times. Simplify the result by performing the multiplications in order, moving 
from left to right. 



Answer: 4 



(-2) 3 = (-2)(-2)(-2) 



(4)(-2) 



—2 as a factor, three times. 
Multiply: (-2) (-2) = 4. 
Multiply: (4) (-2) = -8. 



Thus, (-2) 2 



□ 

In Example 10, note that the product of three negative factors is negative. 
Let's try another example. 



1.1. AN INTROD UCTION TO THE INTEGERS 



11 



EXAMPLE 11. Simplify (-2) 4 . 

Solution: In the exponential expression (— 2) 4 , note that — 2 is the base, while 
4 is the exponent. The exponent tells us to write the base as a factor four 
times. Simplify the result by performing the multiplications in order, moving 
from left to right. 



(-2) 4 = (-2)(-2)(-2)(-2) 
= (4)(-2)(-2) 
= (-8X-2) 
= 16 



-2 as a factor, four times. 



Multiply 
Multiply 
Multiply 



(-2)(-2)=4. 
(4)(-2) = -8. 
(-8)(-2) = 16. 



You Try It! 



Simplify: (-2) 



Thus, (-2) 4 = 16. 



Answer: 



-32 



In Example 11, note that the product of four negative factors is positive. 
Examples 10 and 11 reveal the following pattern. 



□ 



Odd 


or Even 


Exponents. 










1. 


When a 


negative integer 


is raised to 


an even exponent, 


the result is 




positive. 












2. 


When a 


negative integer is raised to an 


odd exponent, 


the 


result is neg- 




ative. 













Graphing Calculator: Negating versus Subtracting 

Consider the view of the lower half of the TI84 graphing calculator in Figure 1.2. 
Note that there are two keys that contain some sort of negative sign, one on 
the bottom row of keys, and another in the last column of keys on the right, 
positioned just above the plus symbol. 



(-) 



and 




Figure 1.2: Lower half of the 
TI-84. 



The first of these buttons is the unary "negation" operator. If you want to 
negate a single (thus the word "unary") number, then this is the key to use. 
For example, enter -3 by pressing the following button sequence. The result is 
shown in Figure 1.3. 



EDH 



ENTER 



12 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



The second button is the binary "subtraction" operator. If you want to subtract 
one number from another number (thus the word "binary" ) , then this is the 
key to use. For example, enter 7-15 by pressing the following button sequence. 
The result is shown in Figure 1.4. 



QH00 



ENTER 




Figure 1.3: Negating a number. 




Figure 1.4: Subtract two numbers. 



Important Point. Do not interchange the roles of the unary negation oper- 


ator and the binary subtraction operator. 


1. To negate a number, use: 


(-) 




2. To subtract one number from another, use: 



If you interchange the roles of these operators, the calculator will respond that 
you've made a "syntax error" (see Figures 1.5 and 1.6). 




ERR: SYNTAX 

IBQuit 

ZSGota 



Figure 1.5: Using the wrong symbol 
for subtraction. 



Figure 1.6: The resulting syntax er- 
ror. 



You Try It! 



Use the graphing calculator 
to evaluate (-225) 3 . 



EXAMPLE 12. Use the TI84 graphing calculator to simplify each of the 
following expressons: 



(a) -717-432 



(b) (232)(-313) 



(c) (-17) 3 



1.1. AN INTROD UCTION TO THE INTEGERS 



13 



Solution: The minus sign in each of these examples looks exactly the same, 
but sometimes it is used as a "negative" sign and sometimes it is used as a 
"subtraction" sign. 

a) The expression —717 — 432 asks us to subtract 432 from "negative" 717. 
Enter the following sequence of keystrokes to produce the result shown in 
the first image in Figure 1.7. 



EDmmmBsss 



ENTER 



Hence, -717-432 



-1149. 



b) The expression (232) (—313) asks us to find the product of 232 and "nega- 
tive" 313. Enter the following sequence of keystrokes to produce the result 
shown in the second image in Figure 1.7. 



HmSHEDHmS 



ENTER 



Hence, (232)(-313) = -72616. 

c) The expression (— 17) 3 asks us to raise "negative" to the third power. Enter 
the following sequence of keystrokes to produce the result shown in the 
third image in Figure 1.7. The "caret" symbol BUB is located just above 
the division key in the rightmost column of the TI84 graphing calculator. 



Eomm 



s 



ENTER 



Hence, (-17) s 



-4913. 



-717-432 



-1149 



232* -313 



-72616 



(-17> A 3 



-4913 



Figure 1.7: Calculations made on the graphing calculator. 



Answer: -11390625 



□ 



14 CHAPTER 1. THE ARITHMETIC OF NUMBERS 



f> t* t* Exercises ■** •** ■** 

In Exercises 1-8, simplify each of the following expressions. 

1. |5|. 5. |2|. 

2. |1|. 6. |8|. 

3. |-2|. 7. |-4|. 

4. |-1|. 8. |-6|. 

In Exercises 9-24, simplify each of the following expressions as much as possible. 

9. -91 + (-147) 17. 37 +(-86) 

10. -23 + (-13) 18. 143 + (-88) 

11.96 + 145 19. 66 +(-85) 

12. 16 + 127 20. 33 + (-41) 

13. -76 + 46 21. 57 + 20 
14.-11 + 21 22.66 + 110 

15. -59 + (-12) 23.-48+127 

16. -40 +(-58) 24.-48 + 92 

In Exercises 25-32, find the difference. 

25. -20- (-10) 29.-77-26 

26. -20 -(-20) 30.-96-92 

27. -62-7 31. -7- (-16) 
28.-82-62 32. -20 -(-5) 

In Exercises 33-40, compute the exact value. 

33. (-8) 6 36. (-4) 6 

34. (-3) 5 37. (-9) 2 

35. (-7) 5 38. (-4) 2 



1.1. AN INTROD UCTION TO THE INTEGERS 



15 



39. (-A) 4 



40. (-5) 4 



In Exercises 41-52, use your graphing calculator to compute the given expression. 

41. -562- 1728 

42. -3125- (-576) 

43. -400- (-8225) 

44. -8176 + 578 

45. (-856)(232) 

46. (-335)(-87) 



47. 


(-815)(-357 


48. 


(753)(-9753) 


49. 


(-18) 3 


50. 


(-16) 4 


51. 


(-13) 5 


52. 


(-15) 6 



£*• £*< £*■ Answers -m •&§ >*& 



1. 5 

3. 2 

5. 2 

7. 4 

9. -238 
11. 241 
13. -30 
15. -71 
17. -49 
19. -19 
21. 77 
23. 79 
25. -10 



27. -69 
29. -103 
31. 9 

33. 262144 
35. -16807 
37. 81 
39. 256 
41. -2290 
43. 7825 
45. -198592 
47. 2916885 
49. -5832 
51. -371293 



16 CHAPTER 1. THE ARITHMETIC OF NUMBERS 

1.2 Order of Operations 

The order in which we evaluate expressions can be ambiguous. Take, for ex- 
ample, the expression —4 + 2-8. If we perform the addition first, then we get 
— 16 as a result (the question mark over the equal sign indicates that the result 
is questionable). 

-4 + 2-8 = -2-8 

? 

= -16. 

On the other hand, if we perform the multiplication first, then we get 12 as a 
result. 

-4 + 2-8 = -4 + 16 
= 12. 

So, what are we to do? 

Of course, grouping symbols would remove the ambiguity. 

Grouping Symbols. Parentheses, brackets, and absolute value bars can be 
used to group parts of an expression. For example: 

3 + 5(9-11) or -2- [-2-5(1-3)] or 6 - 3| - 3 - 4| 

In each case, the rule is "evaluate the expression inside the grouping sym- 
bols first." If the grouping symbols are nested, evaluate the expression in the 
innermost pair of grouping symbols first. 

Thus, if the example above is grouped as follows, we are forced to evaluate the 
expression inside the parentheses first. 

(-4 + 2) • 8 = -2 ■ 8 Parentheses first: -4 + 2 = -2 

= -16 Multiply: -2 ■ 8 = -16 

Another way to avoid ambiguities in evaluating expressions is to establish 
an order in which operations should be performed. The following guidelines 
should always be strictly enforced when evaluating expressions. 



Rules Guiding Order of Operations. When evaluatinj 


5 expressions, pro- 


ceed 


in the following order. 






1. 


Evaluate expressions contained in 


grouping symbols 


first. If grouping 




symbols are nested, evaluate the 


expression in the 


innermost pair of 




grouping symbols first. 







1.2. ORDER OF OPERATIONS 



17 



2. Evaluate all exponents that appear in the expression. 

3. Perform all multiplications and divisions in the order that they appear 
in the expression, moving left to right. 

4. Perform all additions and subtractions in the order that they appear in 
the expression, moving left to right. 



EXAMPLE 1. Simplify: -3-4-8 

Solution: Because of the established Rules Guiding Order of Operations, this 
expression is no longer ambiguous. There are no grouping symbols or exponents 
present, so we immediately go to rule three, evaluate all multiplications and 
divisions in the order that they appear, moving left to right. After that we 
invoke rule four, performing all additions and subtractions in the order that 
they appear, moving left to right. 



You Try It! 



Simplify: -4 + 2 



-3-4- 



-3-32 

-3 + (-32) 
-35 



Multiply first: 4 ■ 8 
Add the opposite. 
Add: -3 +(-32) = 



32 



-35 



Thus, -3-4 



-35. 



Answer: 12 



□ 



Writing Mathematics. When 


simplifying expressions, 


observe the following 


rule to neatly arrange your 


work 












One equal sign per line. 


This 


means that you 


should not arrange 


your work 


horizontally. 














-2-4- (- 


-8) = 


-2-(- 


-32) = - 


-2 + 32 


= 30 




That's three equal signs on 


a sinj 


de line. 


Rather, 


arrange 


your work 


vertically, 


keeping equal signs aligned 


in a i 


:olumn. 










- 


2-4 


(-8) = 


-2-(- 


-32) 












-2 + 32 












30 









18 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: -24/(-3)(2) 



Answer: 16 



EXAMPLE 2. Simplify: 54/(-9)(2) 

Solution: There are no grouping symbols or exponents present, so we imme- 
diately go to rule three, evaluate all multiplications and divisions in the order 
that they appear, moving left to right. 



54/(-9)(2) 



-6(2) 
-12 



Divide first: 54/ (-9) = -6 
Multiply: -6(2) = -12 



Thus, 54/(-9)(2) = -12. 



□ 

Example 2 can be a source of confusion for many readers. Note that multipli- 
cation takes no preference over division, nor does division take preference over 
multiplication. Multiplications and divisions have the same level of preference 
and must be performed in the order that they occur, moving from left to right. 
If not, the wrong answer will be obtained. 

Warning! Here is what happens if you perform the multiplication in Exam- 
ple 2 before the division. 



54/(-9)(2) = 54/(-18) 
= -3 



Multiply: (-9) (2) 
Divide: 54/ (-18) = 



-18 



This is incorrect! Multiplications and divisions must be performed in the 
order that they occur, moving from left to right. 



You Try It! 



Simplify: — 15^ 



EXAMPLE 3. Simplify: (a) (-7) 2 and (b) -7 2 

Solution. Recall that for any integer a, we have (— l)a = —a. Because negat- 
ing is equivalent to multiplying by —1, the Rules Guiding Order of Operations 
require that we address grouping symbols and exponents before negation. 

a) Because of the grouping symbols, we negate first, then square. That is, 

(-7) 2 = (-7)(-7) 
= 49. 

b) There are no grouping symbols in this example. Thus, we must square first, 
then negate. That is, 

-72 



Answer: 



-225 



Thus, (-7) 2 = 49, but -7 2 
(— 7) 2 is different from — 7 2 . 



-(7-7) 
= -49. 

-49. Note: This example demonstrates that 



D 



1.2. ORDER OF OPERATIONS 



19 



Let's try an example that has a mixture of exponents, multiplication, and 
subtraction. 



EXAMPLE 4. Simplify: -3 - 2(-4) 2 

Solution. The Rules Guiding Order of Operations require that we address 
exponents first, then multiplications, then subtractions. 



-4) 2 



-3-2(16) 
-3-32 

-3 +(-32) 
-35 



Exponent first: (—4) = 
Multiply: 2(16) = 32 
Add the opposite. 
Add: -3 +(-32) = -35 



16 



Thus, -3-2(-4) 2 



-35. 



You Try It! 



Simplify: 



4(-2) 2 



Answer: 27 



□ 



Grouping Symbols 

The Rules Guiding Order of Operations require that expressions inside group- 
ing symbols (parentheses, brackets, or curly braces) be evaluated first. 



EXAMPLE 5. Simplify: -2(3 - 4) 2 + 5(1 - 2) 3 

Solution. The Rules Guiding Order of Operations require that we first eval- 
uate the expressions contained inside the grouping symbols. 

-2(3 -4) 2 + 5(1 -2) 3 

= -2(3 + (-4)) 2 + 5(1 + (-2)) 3 Add the opposites. 

= -2(-l) 2 + 5(-l) 3 Parentheses first: 3 + (-4) = -1 

and 1 + (-2) = -1. 



You Try It! 



Simplify: -2-3(-2-3) s 



Evaluate the exponents next, perform the multiplications, then add 

= -2(l) + 5(-l) 

= -2 + (-5) 



Exponents: (—1) =1 
and (-1) 3 = -1. 

Multiply: -2(1) = -2 
and 5(-l) = -5. 

Add: -2 +(-5) = -7 



Thus, -2(3-4)2 + 5(1-2)= 



Answer: 373 



□ 



20 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: 



4-61 



Absolute Value Bars as Grouping Symbols 

Like parentheses and brackets, you must evaluate what is inside them first, 
then take the absolute value of the result. 



EXAMPLE 6. Simplify: -8 - |5 - 11| 

Solution. We must first evaluate what is inside the absolute value bars. 

— 8 — |5 — 11| = -8 - |5 + (-11)| Add the opposite. 

= -8 - | - 6| Add: 5 + (-11) = -6. 

The number —6 is 6 units from zero on the number line. Hence, I — 6| = 6. 



Answer: —10 



Thus, 





= -8-6 


Add: | - 6| = 6. 




= -8 + (-6) 


Add the opposite 




= -14 


Add. 


5- 


- Ill = -14. 





□ 



You Try It! 



Simplify: 

-2-2[-2-2(-2-2)] 



Nested Grouping Symbols 

When grouping symbols are nested, the Rules Guiding Order of Operations tell 
us to evaluate the innermost expressions first. 



EXAMPLE 7. Simplify: -3 - 4[-3 - 4(-3 - 4)] 

Solution. The Rules Guiding Order of Operations require that we first ad- 
dress the expression contained in the innermost grouping symbols. That is, we 
evaluate the expression contained inside the brackets first. 

-3 - 4[-3 - 4(-3 - 4)] = -3 - 4[-3 - 4(-3 + (-4))] Add the opposite. 

= -3 - 4[-3 - 4(-7)] Add: -3 + (-4) = -7 

Next, we evaluate the expression contained inside the brackets. 



-3-4[-3- (-28)] 
-3 - 4[-3 + 28] 
-3 - 4[25] 



Multiply: 4(-7) = -28 
Add the opposite. 
Add: -3 + 28 = 25 



Now we multiply, then subtract. 



1.2. ORDER OF OPERATIONS 



21 



= -3- 100 

= -3 +(-100) 
= -103 

Thus, -3 - 4[-3 - 4(-3 - 4)] 



Multiply: 4{25} = 100 

Add the opposite. 

Add: -3 +(-100) = -103 



-103. 



Answer: 



-14 



□ 



Evaluating Algebraic Expressions 



Variable. A variable is a symbol (usually a letter) that stands for an unknown 
value that may vary. 

Let's add the definition of an algebraic expression. 

Algebraic Expression. When we combine numbers and variables in a valid 
way, using operations such as addition, subtraction, multiplication, division, 
exponentiation, the resulting combination of mathematical symbols is called 
an algebraic expression. 



Thus, 



2a, 



and 



y 



being formed by a combination of numbers, variables, and mathematical oper- 
ators, are valid algebraic expressions. 

An algebraic expression must be well-formed. For example, 

2 + -5a; 

is not a valid expression because there is no term following the plus sign (it is 
not valid to write H — with nothing between these operators). Similarly, 

2 + 3(2 

is not well-formed because parentheses are not balanced. 

In this section we will evaluate algebraic expressions for given values of the 
variables contained in the expressions. Here are some simple tips to help you 
be successful. 



Tips for Evaluating Algebraic Expressions. 



22 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



1. Replace all occurrences of variables in the expression with open paren- 
theses. Leave room between the parentheses to substitute the given value 
of the variable. 

2. Substitute the given values of variables in the open parentheses prepared 
in the first step. 

3. Evaluate the resulting expression according to the Rules Guiding Order 
of Operations. 



You Try It! 



If X: 

evaluate x 3 - 



2 and y = — 1 , 

y 3 



EXAMPLE 8. Evaluate the expression x 2 — 2xy + y 2 at x = —3 and y = 2. 

Solution. Following Tips for Evaluating Algebraic Expressions, first replace all 
occurrences of variables in the expression x 2 — Ixy + y 2 with open parentheses. 
Next, substitute the given values of variables (—3 for x and 2 for y) in the open 
parentheses. 

x 2 -2xy + y 2 = ( ) 2 -2( )( ) + ( ) 2 
= (-3) 2 -2(-3)(2) + (2) 2 

Finally, follow the Rules Guiding Order of Operations to evaluate the resulting 
expression. 



2xy + y 

-( ) 2 "2( )() + () 2 
: (-3) 2 -2(-3)(2) + (2) 2 
= 9-2(-3)(2)+4 
= 9-(-6)(2) + 4 
9- (-12) + 4 
: 9+ 12 + 4 

:25 



Original expression. 

Replace variables with parentheses. 

Substitute —3 for x and 2 for y. 

Evaluate exponents first. 

Left to right, multiply: 2(— 3) = —6 

Left to right, multiply: (-6) (2) = -12 

Add the opposite. 

Add. 



Answer: 



Thus, if x = —3 and y = 2, then x 2 — 2xy + y 2 



25. 



□ 



Evaluating Fractions 

If a fraction bar is present, evaluate the numerator and denominator separately 
according to the Rules Guiding Order of Operations, then perform the division 
in the final step. 



1.2. ORDER OF OPERATIONS 



23 



You Try It! 



EXAMPLE 9. Evaluate the expression 

ad — be 
a + b 

at a = 5, b = —3, c = 2, and d = —4. 

Solution. Following Tips for Evaluating Algebraic Expressions, first replace 
all occurrences of variables in the expression (ad — be)/ (a + b) with open paren- 
theses. Next, substitute the given values of variables (5 for a, —3 for b, 2 for c, 
and —4 for d) in the open parentheses. 

ad-be _ {){ )-( )( ) 
a+b ()+( ) 

(5)(-4) - (-3)(2) 
(5) + (-3) 

Finally, follow the Rules Guiding Order of Operations to evaluate the result- 
ing expression. Note that we evaluate the expressions in the numerator and 
denominator separately, then divide. 



ad-bc ( )( )-( )( ) 



a + b 




( ) + 


( 


) 




= 


(5)(-4)- 


-(" 


-3)(2) 




(5) + 


(" 


3) 






-20- (- 


-6) 








2 










-20 + 6 










2 
-14 










2 








= 


-7 






Thus, if 


« 


= 5, b = - 


-3, 


c = 2, 



Replace variables with parentheses. 

Substitute: 5 for a, —3 for 6, 2 for c, —4 for d 

Numerator: (5)(-4) = -20, (-3)(2) = -6 
Denominator: 5 + (— 3) = 2 

Numerator: Add the opposite 

Numerator: —20 + 6 = —14 
Divide. 

2, and d = -4, then (ad - be)/ (a + b) = -7. 



If a = 
and d 



-7, b = —3, c = —15, 
- — 14, evaluate: 



Answer: 



□ 



Using the Graphing Calculator 

The graphing calculator is a splendid tool for evaluating algebraic expressions, 
particularly when the numbers involved are large. 



24 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Use the graphing calculator 
to evaluate 

-2-2[-2-2(-2-2)]. 



EXAMPLE 10. Use the graphing calculator to simplify the following expres- 
sion. 

-213 - 35[-18 - 211(15 - 223)] 

Solution. The first difficulty with this expression is the fact that the graphing 
calculator does not have a bracket symbol for the purposes of grouping. The 
calculator has only parentheses for grouping. So we first convert our expression 
to the following: 

-213 - 35(-18 - 211(15 - 223)) 

Note that brackets and parentheses are completely interchangeable. 

The next difficulty is determining which of the minus signs are negation 
symbols and which are subtraction symbols. If the minus sign does not appear 
between two numbers, it is a negation symbol. If the minus sign does appear 
between two numbers, it is a subtraction symbol. Hence, we enter the following 
keystrokes on our calculator. The result is shown in Figure 1.8. 



ED0QSB00DBEDS0BH 

QQBiQEI000 



ENTER 



Answer: —14 



-2 13-35* (-18-211 
*U5-223>> 

-1535663 



Figure 1.8: Calculating -213 - 35[-18 - 211(15 - 223)]. 



Thus, -213 -35[-18- 211(15 -223)] = -1,535,663. 



□ 



You Try It! 



Use the graphing calculator 
to evaluate 

10 + 10 
10 + 10 



EXAMPLE 11. Use the graphing calculator to evaluate 

5 + 5 
5 + 5 



1.2. ORDER OF OPERATIONS 



2o 



5 + 5 


10 


5 + 5 


" 10 




= 1 



Solution. You might ask "Why do we need a calculator to evaluate this 
exceedingly simple expression?" After all, it's very easy to compute. 

Simplify numerator and denominator. 
Divide: 10/10 = 1. 

Well, let's enter the expression 5+5/5+5 in the calculator and see how well 
we understand the Rules Guiding Order of Operations (see first image in 
Figure 1.9). Whoa! How did the calculator get 11? The answer is supposed to 
be 1! 

Let's slow down and apply the Rules Guiding Order of Operations to the 
expression 5+5/5+5. 

5 + 5/5 + 5 = 5+ - + 5 
5 

=5+1+5 

o 

= 11 Add: 5 + 1 + 5 = 

Aha! That's how the calculator got 11. 

5 + 5/5 + 5 is equivalent to 

Let's change the order of evaluation by using grouping symbols. Note that: 



ivide first. 




ivide: — = 
5 


1. 


dd: 5 + 1 + 5 


-1 


+ 5 



(5 + 5)/(5 + 5) = 10/10 
= 1 



Parentheses first. 
Divide: 10/10= 1. 



That is: 



(5 + 5)/(5 + 5) is equivalent to 



5 + 5 



Enter (5+5) (5+5) and press the ENTER key to produce the output shown in 
the second image in Figure 1.9. 



5+5/5+5 



11 



(5+5VC5+5) 



Figure 1.9: Calculating 



5 + 5 
5 + 5' 



Answer: 1 



□ 



26 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Use the graphing calculator 
to evaluate |o — b\ at 
a = -312 and b = -875. 




9*a ;'0 



Figure 1.10: Upper half of the 
TI84. 



The graphing calculator has memory locations available for "storing" values. 
They are lettered A-Z and appear on the calculator case, in alphabetic order 
as you move from left to right and down the keyboard. Storing values in 
these memory locations is an efficient way to evaluate algebraic expressions 
containing variables. Use the key to access these memory locations. 



EXAMPLE 12. Use the graphing calculator to evaluate \a\ 
and b = -875. 



at a 



-312 



Solution. First store —312 in the variable A with the following keystrokes. 
To select the letter A, press the ALPHA key, then the MATH key, located in 
the upper left-hand corner of the calculator (see Figure 1.10). 



EDHQE 



ALPHA 



ENTER 



Next, store —875 in the variable B with the following keystrokes. To select the 
letter B, press the ALPHA key, then the APPS key. 



1000 



ALPHA 



The results of these keystrokes are shown in the first image in Figure 1.11. 

Now we need to enter the expression \a\ — \b\. The absolute value function 
is located in the MATH menu. When you press the MATH key, you'll notice 
submenus MATH, NUM, CPX, and PRB across the top row of the MATH 
menu. Use the right-arrow key to select the NUM submenu (see the second 
image in Figure 1.11). Note that abs( is the first entry on this menu. This 
is the absolute value function needed for this example. Enter the expression 
abs(A)-abs(B) as shown in the third image in Figure 1.11. Use the ALPHA 
key as described above to enter the variables A and B and close the parentheses 
using the right parentheses key from the keyboard. Press the ENTER key to 
evaluate your expression. 



-312-»FI 
-875-»B 



-312 
-S75 



MATH [Bill CPX PRE 

iHabs< 

2: round ( 

3: iPartC 

4: f PartC 

5: int( 

6: min< 

7-1-maxt 




Figure 1.11: Evaluate \a\ — \b\ at a = —312 and b 



-875. 



Answer: 563 



Thus, \a\ 



-563. 



□ 



10. 


-(-30) 


11. 


-3 5 


12. 


-3 2 


13. 


48 -r 4(6) 


14. 


96 -=-6(4) 


15. 


-52-8(-8 


16. 


-8-7(-3) 


17. 


("2) 4 


18. 


("4) 4 



1.2. ORDER OF OPERATIONS 27 



».**.**• Exercises ■** * «•* 



In Exercises 1-18, simplify the given expression. 

1. -12 + 6(-4) 

2. 11 + 11(7) 

3. -(-2) 5 

4. -(-5) 3 

5. — | — 40| 

6. -|-42| 

7. -24/(-6)(-l) 

8. 45/(-3)(3) 

9. -(-50) 



In Exercises 19-42, simplify the given expression. 

19. 9-3(2) 2 31.-6-5(4-6) 

20. -4-4(2) 2 32. -5-5(-7-7) 

21. 17— 10|13 — 14| 33. 9 + (9- 6) 3 - 5 

22. 18 — 3| — 20 — 5| 34. 12 + (8 -3) 3 -6 

23. -4 + 5(-4) 3 35. -5 + 3(4) 2 

24. 3 + 3(-4) 3 36.2 + 3(2) 2 

25. 8-5(-l-6) 37. 8- (5 - 2) 3 + 6 

26. 8-4(-5-5) 38. 9- (12- ll) 2 +4 

27. (10 - 8) 2 - (7 - 5) 3 39. |6 - 15| - | - 17 - 11| 

28. (8 - 10) 2 - (4 - 5) 3 40. | - 18 - 19| - | - 3 - 12| 

29. 6 - 9(6 - 4(9 - 7)) 41. 5 - 5(5 - 6(6 - 4)) 

30. 4 - 3(3 - 5(7 - 2)) 42. 4 - 6(4 - 7(8 - 5)) 



28 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



In Exercises 43-58, evaluate the expression at the given values of x and y 

43. 4a; 2 + 3xy + Ay 2 at x = -3 and y = 

44. 3a; 2 - 3xy + 2y 2 at x = 4 and y = -3 

45. -8x + 9 at x = -9 

46. -12a; + 10 at x = 2 

47. — 5x 2 + 2xy — 4y 2 at a; = 5 and y = 

48. 3a; 2 + 3xy - by 2 at x = and y = 3 

49. 3a; 2 + 3a; - 4 at a; = 5 

50. 2a; 2 + 6x - 5 at a; = 6 



51. -2a; 2 + 2y 2 at a; = 1 and y = -2 

52. -5a; 2 + 5y 2 at x = -4 and y = 

53. -3a; 2 - 6a; + 3 at x = 2 

54. -7a; 2 + 9x + 5 at x = -7 

55. — 6a; — 1 at a; = 1 

56. 10a; + 7 at x = 9 

57. 3a; 2 - 2y 2 at a: = -3 and y = -2 

58. -3a; 2 + 2y 2 at x = 2 and y = 2 



59. Evaluate 

a 2 + 6 2 

a + 6 
at a = 27 and 6 = -30. 

60. Evaluate 

a 2 + b 2 

a + b 
at a = —63 and b = 77. 



61. Evaluate 



a + b 



c— d 
at a = -42, b = 25, c = 26, and d = 43. 



62. Evaluate 



a + 6 



c — d 
at a = 38, b = 42, c = 10, and d = 50. 

63. Evaluate 

a — b 



cd 



at a = —7, b = 48, c = 5, and d = 11. 
64. Evaluate 



cd 



at a = -46, 6 = 46, c = 23, and d = 2. 



65. Evaluate the expressions a 2 + b 2 and (a + 6) 2 at a = 3 and 6 = 4. Do the expressions produce the 
same results? 

66. Evaluate the expressions a 2 b 2 and (ab) 2 at a = 3 and 6 = 4. Do the expressions produce the same 
results? 

67. Evaluate the expressions |a||6| and \ab\ at a = —3 and 6 = 5. Do the expressions produce the 
same results? 

68. Evaluate the expressions \a\ + \b\ and \a + b\ at a = —3 and 6 = 5. Do the expressions produce 
the same results? 



1.2. ORDER OF OPERATIONS 



2!) 



In Exercises 69-72, use a graphing calculator to evaluate the given expression. 
69. -236-324(-576 + 57) — 270-900 



70. -443 + 27(-414-22) 



71. 

72. 



300 - 174 

3000 - 952 

144 - 400 



73. Use a graphing calculator to evaluate the expression 



a 



a 



at a = —93 and b = 84 by first storing 



-93 in the variable A and 84 in the variable B, then entering the expression (A~2+B~2)/ (A+B) . 

.2 , ia 



74. Use a graphing calculator to evaluate the expression 



a 



a + b 



at a = — 76 and b = 77 by first storing 



-76 in the variable A and 77 in the variable B, then entering the expression (A~2+B~2)/ (A+B) . 



75. The formula 



9 



C + 32 



will change a Celsius temperature to a 
Fahrenheit temperature. Given that the 
Celsius temperature is C = 60° C, find the 
equivalent Fahrenheit temperature. 

76. The surface area of a cardboard box is 
given by the formula 

S = 2WH + 2LH + 2LW, 

where W and L are the width and length 
of the base of the box and H is its height. 
If W = 2 centimeters, L = 8 centimeters, 
and H = 2 centimeters, find the surface 
area of the box. 

77. The kinetic energy (in joules) of an object 
having mass m (in kilograms) and veloc- 
ity v (in meters per second) is given by 



the formula 

K = —mv . 
2 

Given that the mass of the object is m = 7 
kilograms and its velocity is v = 50 meters 
per second, calculate the kinetic energy of 
the object. 

78. The area of a trapezoid is given by the 
formula 



A = -(b 1 + b 2 ) h, 



where b± and 62 are the lengths of the 
parallel bases and h is the height of the 
trapezoid. If the lengths of the bases are 
21 yards and 11 yards, respectively, and if 
the height is 22 yards, find the area of the 
trapezoid. 



30 CHAPTER 1. THE ARITHMETIC OF NUMBERS 

s*- j*- £*■ Answers <#$ >#s >*$ 

1. -36 41. 40 

3. 32 43. 36 

5. -40 45. 81 

7 - - 4 47. -125 

9. 50 49 86 
11. -243 



13. 72 
15. 12 
17. 16 
19. -3 
21. 7 
23. -324 
25. -27 
27. -4 
29. 24 
31. 4 
33. 31 
35. 43 
37. -13 
39. -19 



51. 6 

53. -21 

55. -7 

57. 19 

59. -543 

61. 1 

63. -1 

65. No. 

67. Yes. 

69. 167920 

71. -5 

73. -8 

75. 140° F 

77. 8750 joules 



1.3. THE RATIONAL NUMBERS 



31 



1.3 The Rational Numbers 

We begin with the definition of a rational number. 



Rational Numbers. Any number that can be expressed in the form 
where p and q are integers, q 7^ 0, is called a rational number. The letter 
used to represent the set of rational numbers. That is: 


p/q, 

Qis 




Q-- 


= < — : p and 
U 


q are integers, q 7^ > 







Because —2/3, 4/5, and 123/(— 12) have the form p/q, where p and q are 
integers, each is an example of a rational number. If you think you hear 
the word "fraction" when we say "rational number," you are correct in your 
thinking. Any number that can be expressed as a fraction, where the numerator 
and denominator are integers, is a rational number. 

Every integer is also a rational number. Take, for example, the integer 
— 12. There are a number of ways we can express —12 as a fraction with 
integer numerator and denominator, —12/1, 24/ (—2), and —36/3 being a few. 

Reducing Fractions to Lowest Terms 

First, we define what is meant by the greatest common divisor of two integers. 

The Greatest Common Divisor. Given two integers a and b, the greatest 
common divisor of a and 6 is the largest integer that divides evenly (with no 
remainder) into both a and b. The notation GCD(a, b) is used to represent the 
greatest common divisor of a and b. 



For example, GCD(12, 18) = 6, GCD(32, 40) = 8, and GCD(18, 27) = 9. 
We can now state when a fraction is reduced to lowest terms. 



Lowest Terms. A fraction a/b is said to be reduced to lowest terms if and 
only if GCD(a,6) = 1. 



A common technique used to reduce a fraction to lowest terms is to divide both 
numerator and denominator by their greatest common denominator. 



EXAMPLE 1. Reduce 8/12 to lowest terms. 



You Try It! 



Reduce: -48/60 



32 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: —4/5 



Solution: Note that GCD(8, 12) = 4. Divide both numerator and denomina- 
tor by 4. 



-r4 



12 12 -=-4 

_ 2 
~ 3 

Thus, 8/12 = 2/3. 



Divide numerator and denominator 
by GCD(8,12) = 4. 

Simplify numerator and denominator. 



□ 



Recall the definition of a prime number. 

Prime Number. A natural number greater than one is prime if and only if 
its only divisors are one and itself. 



You Try It! 



Reduce 18/24 to lowest 
terms. 



For example, 7 is prime (its only divisors are 1 and 7), but 14 is not (its divisors 
are 1, 2, 7, and 14). In like fashion, 2, 3, and 5 are prime, but 6, 15, and 21 
are not prime. 



EXAMPLE 2. Reduce 10/40 to lowest terms. 

Solution: Note that GCD(10,40) = 10. Divide numerator and denominator 
by 10. 



Divide numerator and denominator 
by GCD(10,40) = 10. 

Simplify numerator and denominator. 



10 


10- 


-10 


40 


40- 

1 

~ 4 


-10 



Alternate solution: Use factor trees to express both numerator and denom- 
inator as a product of prime factors. 



10 



>D 



40 
/ \ 

4 10 



2 2 



Hence, 10 = 2 ■ 5 and 40 = 2 • 2 • 2 • 5. Now, to reduce 10/40 to lowest terms, 
replace the numerator and denominator with their prime factorizations, then 
cancel factors that are in common to both numerator and denominator. 



1.3. THE RATIONAL NUMBERS 



:W 



10 

40 



2-5 



2-2-2-5 
1 



Prime factor numerator and denominator. 

Cancel common factors. 

Simplify numerator and denominator. 



When we cancel a 2 from both the numerator and denominator, we're actually 
dividing both numerator and denominator by 2. A similar statement can be 
made about canceling the 5. Canceling both 2 and a 5 is equivalent to dividing 
both numerator and denominator by 10. This explains the 1 in the numerator 
when all factors cancel. 



Answer: 3/4 



Example 2 demonstrates an important point. 

When all factors cancel. When all of the factors cancel in either numerator 
or denominator, the resulting numerator or denominator is equal to one. 



□ 



Multiplying Fractions 

First, the definition. 



Multiplication of Fractions. 


If a/b and c/d are two fractions, then their 


product is defined as follows: 






a c ac 
1"d~ bd 



Thus, to find the product of a/b and c/d, simply multiply numerators and 
multiply denominators. For example: 



1 3 

2 ' 4 



and 



2 7 
5 ' 3 



14 
15 



and 



5_ 
48 



Like integer multiplication, like signs yield a positive answer, unlike signs yield 
a negative answer. 

Of course, when necessary, remember to reduce your answer to lowest terms. 



EXAMPLE 3. Simplify: 



14 10 
"20 ' 21 



You Try It! 



Simplify: 



27 
20 



34 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Solution: Multiply numerators and denominators, then reduce to lowest terms. 

Multiply numerators 
and denominators. 

Prime factor. 

Cancel common factors. 

Simplify. 

Note that when all the factors cancel from the numerator, you are left with a 
Answer: 6/5 1. Thus, (-14/20) • (10/21) = -1/3. 

□ 



14 10 
20 ' 21 ~ 


140 
~420 

2-2-5-7 




2-2-3-5-7 


= 


1 
~3 



Cancellation Rule. When multiplying fractions, cancel common factors ac- 
cording to the following rule: "Cancel a factor in a numerator for an identical 
factor in a denominator." 



You Try It! 



Simplify: 



_6_ 

45 



35 
14 



The rule is "cancel something on the top for something on the bottom." 

Thus, an alternate approach to multiplying fractions is to factor numerators 
and denominators in place, then cancel a factor in a numerator for an identical 
factor in a denominator. 



EXAMPLE 4. Simplify: 



15 



14 
~9~ 



Solution: Factor numerators and denominators in place, then cancel common 
factors in the numerators for common factors in the denominators. 



15 



t) 


3-5 
2-2-2 


/ 2-7 




2-5 
2-2-2 


( 2-7 
V #'3 




35 
~12 





Factor numerators 
and denominators. 

Cancel a factor in a 
numerator for a common, 
factor in a denominator. 

Multiply numerators and. 
denominators. 



Answer: 1/3 



Note that unlike signs yield a negative product. Thus, (15/8) ■ (—14/9) 
-35/12. 



□ 



1.3. THE RATIONAL NUMBERS 



35 



Dividing Fractions 

Every nonzero rational number has was it called a multiplicative inverse or 
reciprocal. 

The Reciprocal. If a is any nonzero rational number, then l/a is called the 
multiplicative inverse or reciprocal of a, and: 

1 

a- - = 1 

a 



Note that: 



and 



3 5 
5 ' 3 



and 



1. 



Thus, the reciprocal of 2 is 1/2, the reciprocal of 3/5 is 5/3, and the reciprocal 
of —4/7 is —7/4. Note that to find the reciprocal of a number, simply invert 
the number (flip it upside down). 

Now we can define the quotient of two fractions. 

Division of Fractions. If a/b and c/d are two fractions, then their quotient 

is defined as follows: 

a _ c ad 

b d b c 

That is, dividing by c/d is the same as multiplying by the reciprocal d/c. 



The above definition of division is summarized by the phrase "invert and mul- 
tiply." 







You Try It! 










35 / 
EXAMPLE 5. Simplify: -=- ( - 


1CA 

12/ Simplif 


4 27 

v: '-. 

Q 81 




Solution: Invert and multiply, then factor in place and cancel common factors 


in a numerator for common factors in 


a denominator. 


35 . / 10\ 35 / 12\ 

~2i ~ v 12/ ~~ ~2i ' v Toy 


Invert and multiply. 


5-7 ( 2-2-3" 
3 • 7 V 2 • 5 , 


I Prime factor. 


$•1 ( t-1-P 
t-1 \ %•$ , 


1 Cancel common factors. 


2 
~ 1 


Multiply numerators and denominators. 


= 2 


Simplify. 







36 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: 



-4/3 



Note that when all the factors in a denominator cancel, a 1 remains. Thus, 
(—35/21) -T- (—10/12) = 2. Note also that like signs yield a positive result. 

□ 



Adding Fractions 

First the definition. 

Addition of Fractions. If two fractions have a denominator in common, 
add the numerators and place the result over the common denominator. In 
symbols: 

a b a + b 



For example: 

3 7 _ 4 
~5 + 5 ~ 5 



and 



11 
T 



and 



If the fractions do not posses a common denominator, first create equivalent 
fractions with a least common denominator, then add according to the rule 
above. 

Least Common Denominator. If the fractions a/b and c/d do not share 
a common denominator, the least common denominator for b and d, written 
LCD(&,eQ, is defined as the smallest number divisible by both b and d. 



You Try It! 



Simplify: 1 — 

1 y 6 9 



EXAMPLE 6. Simplify: 



5_ 
12 



Solution: The least common denominator in this case is the smallest number 
divisible by both 8 and 12. In this case, LCD(8, 12) = 24. We first need to 
make equivalent fractions with a common denominator of 24. 



Answer: 



-13/18 



8 + 12 



3 3 5 

~8 ' 3 + 12 


2 
2 


9 10 

~24 + 24 
1 




24 





Make equivalent fraction with 
a common denominator of 24. 

Multiply numerators and denominators. 



Add: -9 + 10 = 1. 



Note how we add the numerators in the last step, placing the result over the 
common denominator. Thus, —3/8 + 5/12 = 1/24. 



□ 



1.3. THE RATIONAL NUMBERS 



37 



Order of Operations 

Rational numbers obey the same Rules Guiding Order of Operations as do the 
integers. 



Rules Guiding Order of Operations. When evaluating expressions, pro- 
ceed in the following order. 

1. Evaluate expressions contained in grouping symbols first. If grouping 
symbols are nested, evaluate the expression in the innermost pair of 
grouping symbols first. 

2. Evaluate all exponents that appear in the expression. 

3. Perform all multiplications and divisions in the order that they appear 
in the expression, moving left to right. 

4. Perform all additions and subtractions in the order that they appear in 
the expression, moving left to right. 



You Try It! 



EXAMPLE 7. Given x = 2/3, y = -3/5, and z = 10/9, evaluate xy + yz. 



Solution: Following Tips for Evaluating Algebraic Expressions, first replace 
all occurrences of variables in the expression xy + yz with open parentheses. 
Next, substitute the given values of variables (2/3 for x, —3/5 for y, and 10/9 
for z) in the open parentheses. 



Given a = -1/2, b= 2/3, 
and c = —3/4, evaluate the 
expression a + be and 
simplify the result. 



xy + yz 



10 



Replace variables with parentheses. 



Substitute: 2/3 for x, —3/5 
for y, and 10/9 for z. 



38 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: 



Use the Rules Guiding Order of Operations to simplify. 


6 / 30\ 
~~15 + V 45/ 


Multiply. 


-H-l) 


Reduce. 


2 3 / 2 5\ 
~~5 ' 3 + V 3 ' 5/ 


Make equival 


6 / 10\ 

= -15 + ( v -15J 


least common 


16 

~ ~15 


Add. 



Thus, if x = 2/3, y = —3/5, and z = 10/9, then xy + yz = —16/15 



□ 



You Try It! 



Simplify: 



-1/3) 4 



EXAMPLE 8. Given x = -3/5, evaluate -x 3 . 

Solution: First, replace each occurrence of the variable x with open paren- 
theses, then substitute —3/5 for x. 



Answer: 1/81 



( r 



3 

~5 

3 

5 

27 
125 



27 
125 



Replace x with open parentheses. 
Substitute —3/5 for x. 

Write —3/5 as a factor 
three times. 

The product of three negative 

fractions is negative. Multiply 
numerators and denominators. 

The opposite of -27/125 is 27/125. 



Hence, 



27/125, given x = -3/5. 



□ 



You Try It! 



Given x = —3/4 and 

y = —4/5, evaluate x 2 — y 2 . 



EXAMPLE 9. Given a = -4/3 and b = -3/2, evaluate a 2 + 2ab - 3b 2 . 

Solution: Following Tips for Evaluating Algebraic Expressions, first replace all 
occurrences of variables in the expression a 2 + lab— 3& 2 with open parentheses. 



1.3. THE RATIONAL NUMBERS 



39 



Next, substitute the given values of variables (—4/3 for a and —3/2 for b) in 
the open parentheses. 



a 2 + 2ab - 36 2 



Next, evaluate the exponents: (-4/3) 2 = 16/9 and (-3/2) 2 = 9/4. 



16 2 



3 /9 

i u 



Next, perform the multiplications and reduce. 

_ 16 24 27 

~ ~9 + ~6 ~ T 

16 27 

= h4 

9 4 

Make equivalent fractions with a common denominator, then add. 

16 4 36 27 9 

~ T ' 4 + ' 36 ~ T ' 9 
_ 64 144 243 

~ 36 + "36" ~ lifT 
35 

~~36 



Thus, if a = -4/3 and b = -3/2, then a 2 + 2al 



3b 2 



-35/36 



Answer: 



-31/400 



□ 



Fractions on the Graphing Calculator 

We must always remember that the graphing calculator is an "approximating 
machine." In a small number of situations, it is capable of giving an exact 
answer, but for most calculations, the best we can hope for is an approximate 
answer. 

However, the calculator gives accurate results for operations involving frac- 
tions, as long as we don't use fractions with denominators that are too large 
for the calculator to respond with an exact answer. 



40 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify using the graphing 
calculator: 

4 8 
"5 + 3 



EXAMPLE 10. Use the graphing calculator to simplify each of the following 
expressions: 

1 
2 
Solution: We enter each expression in turn. 



(a) - 
V ; 3 



y ' 3 7 



M 3 . 1 

(C) 5^3 



a) The Rules Guiding Order of Operations tell us that we must perform divi- 
sions before additions. Thus, the expression 2/3 + 1/2 is equivalent to: 



2/3 



2 1 

1/2= - + - 
1 3 2 

4 3 

~ 6 + 6 

_ 7 

~ 6 



Divide first. 

Equivalent fractions with LCD. 

Add. 



Enter the expression 2/3+1/2 on your calculator, then press the ENTER 
key. The result is shown in the first image in Figure 1.12. Next, press the 
MATH button, then select l:^Frac (see the second image in Figure 1.12) 
and press the ENTER key again. Note that the result shown in the third 
image in Figure 1.12 matches the correct answer of 7/6 found above. 



2/3+1/2 




BJJ1I1 HUM CPtt PRE 




2/3+1/2 


1.166666667 




ija^Frac 




1.166666667 


I 




2: ►Dec 

3:^ 

4:*J-< 

5: x j 

6ifMin<: 

7-1-fMaxC 




flns^Frac 

7/6 

I 



Figure 1.12: Calculating 2/3+1/2. 



b) The Rules Guiding Order of Operations tell us that there is no preference 
for division over multiplication, or vice-versa. We must perform divisions 
and multiplications as they occur, moving from left to right. Hence: 



/3x 5/7 = 


2xV7 


2 
Divide: 2/3 = - 




^ 


2 10 
Multiply: - x 5 = — 




10 1 

T x 7 


Invert and multiply. 




10 
21 


10 1 10 

Multiply: — x - = — 

y - y 3 7 21 



1.3. THE RATIONAL NUMBERS 



41 



This is precisely the same result we get when we perform the following 
calculation. 

10 

21 
Hence: 

2/3x5/7 is equivalent to 



2 5 
- x - 

3 7 



Multiply numerators and denominators. 



2 5 
- x - 

3 7 



Enter the expression 2/3x5/7 on your calculator, then press the ENTER 
key. The result is shown in the first image in Figure 1.13. Next, press the 
MATH button, then select 1 : ►Frac (see the second image in Figure 1.13) 
and press the ENTER key again. Note that the result shown in the third 
image in Figure 1.13 matches the correct answer of 10/21 found above. 



2/3*5/7 




araii hum cpx pre 




2/3*5/7 


.4761904762 




Ija^Frac 




.4761904762 


I 




2: ►Dec 

3:^ 

4:*J-< 

5: x j 

fcifMinC 

7-lfMaxC 




fins ► Frac 

10/21 



Figure 1.13: Calculating 2/3 x 1/2. 



This example demonstrates that we need a constant reminder of the Rules 
Guiding Order of Operations. We know we need to invert and multiply in 
this situation. 



3 3 

- x - 
5 1 

9 

5 



Invert and multiply. 

Multiply numerators and denominators. 



So, the correct answer is 9/5. 

Enter the expression 3/5/1/3 on your calculator, then press the ENTER 
key. Select l:^Frac from the MATH menu and press the ENTER key 
again. Note that the result in the first image in Figure 1.14 does not 
match the correct answer of 9/5 found above. What have we done wrong? 

If we follow the Rules Guiding Order of Operations exactly, then: 



3/5/1/3 =g/l/3 

-> 

_ 3 1 

~ 5 X 3 
1 



Divide: 3/5 = 

3 
Divide: -/l 



3 

5 
3 

' 5 



Invert and multiply. 

3 11 

Multiply: - x - = - 
5 3 5 



42 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: 28/15 



This explains the answer found in the first image in Figure 1.14. However, 
it also show that: 



3 ; 1 
5^3 



3/5/1/3 is not equivalent to 

We can cure the problem by using grouping symbols. 

Parentheses first. 

/ is equivalent to -j- 
u o 

Hence: 



(3/5)/(l/3)= |/| 

_ 3 _ 1 

~ 5 ~ 3 



(3/5)/(l/3) is equivalent to 



3 , 1 
5^3 

Enter the expression (3/5)/ (1/3) on your calculator, then press the EN- 
TER key. Select 1 : ►Frac from the MATH menu and press the ENTER 
key again. Note that the result in the second image in Figure 1.14 matches 
the correct answer of 9/5. 



3/5/1/3 
fins ► Frac 



.2 
1/5 



(3/5)/(l/3) 
fins ► Frac 



l.S 
9/5 



Figure 1.14: Calculating (3/5)/(l/3). 



□ 



1.3. THE RATIONAL NUMBERS 43 



**.**.**. Exercises -*s -« ■*? 

In Exercises 1-6, reduce the given fraction to lowest terms by dividing numerator and denominator by 
the their greatest common divisor. 

20 36 

1. — 4. — 

50 14 

2. H 5* 

38 45 

10 21 

3. — 6. — 

48 36 



In Exercises 7-13, reduce the given fraction to lowest terms by prime factoring both numerator and 
denominator and cenceling common factors. 

~ 153 ,~ 171 

7. 10. 

170 144 

198 159 

8. 11. 

144 106 

188 140 

9. 12. 

141 133 



In Exercises 13-18, for each of the following problems, multiply numerators and denominators, then 
prime factor and cancel to reduce your answer to lowest terms. 



13 

18/2 

14. — ■ — 

16 V 5 



15.-H.f_i? 
4 V 13 



16. 


3 
~2 ' 


H? 


17. 


16 

~~8~ ' 


19 


18. 


14 


7 
17 



In Exercises 19-24, for each of the following problems, first prime factor all numerators and denomina- 
tors, then cancel. After canceling, multiply numerators and denominators. 



5 / 12\ 36 / 21 
19. 20. 

6 V 49/ 17 V 46 



44 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



21. 



22. 



21 12 
10 ' 55 

49 52 
13 ' 51 



23. 


55 

29 ' 


(-s 


24. 


7 
13 ' 


/ 55 

^~49 



In Exercises 25-30, divide. Be sure your answer is reduced to lowest terms. 



75 


50 


/ 5 




39 


V 58 


76. 


25 


{-!) 


77. 


60 . 
~17^ 


34 
"31 



?8 


27 . 


45 




28 


23 


79. 


7 
10 


{ 28 


30. 


4 
~13 ~ 


/ 48 
" V 35 



In Exercises 31-38, add or subtract the fractions, as indicated, and simplify your result. 



31.4 



37. 



33. — 

9 



34. 



35 "4 



36. — 

2 



37. 

9 5 

4 1 
38. 

7 3 



In Exercises 39-52, simplify the expression. 



39. 



40. 



41. 



5 2 
2 ~ 5 

7 1 

6 ~ 2 

2 



«■ 1 1 



1 

2 

iV 5 



44. 



45. 



46. 



47. 



4MB 



43. 



48. 



1.3. THE RATIONAL NUMBERS 



45 



49. 



50. 



6 



2 
5 5 

3 5 



(i 



51. 



52. 



In Exercises 53-70, evaluate the expression at the given values. 



53. xy 



z at x 



-1/2, y = -1/3, and 



z = 5/2 

54. xy—z 2 at x = —1/3, y = 5/6, and z = 1/3 

55. -5a; 2 + 2y 2 at x = 3/4 and y = -1/2. 

56. -2.x 2 + Ay 2 at x = 4/3 and y = -3/2. 

57. 2a; 2 - 2xy - 3y 2 at x = 3/2 and y = -3/4. 

58. 5a; 2 - 4xy- 3y 2 at a; = 1/5 and y = -4/3. 

59. x + yz at x = —1/3, y = 1/6, and z = 2/5. 

60. x + yz a,t x = 1/2, y = 7/4, and 2 = 2/3. 

61. ab+bc at a = -4/7, 6 = 7/5, and c = -5/2 



62. ab+bc at a = -8/5, 6 = 7/2, and c = -9/7 

63. x 3 at a; = -1/2 

64. a; 2 at a; = -3/2 

65. x—yz at a; = —8/5, y = 1/3, and z = —8/5 

66. x — yz at x = 2/3, y = 2/9, and z = —3/5 

67. -a; 2 at a; = -8/3 

68. -x 4 at x = -9/7 

69. a; 2 + yz at a; = 7/2, y = -5/4, and 
z = -5/3 

70. a; 2 + yz at x = 1/2, y = 7/8, and z = -5/9 



71. a + b/c + d is equivalent to which of the 
following mathematical expressions? 

a + b 



(a) a H h d 

c 

(c) + d 



(b) 



c + d 



(d)a- 



c c + d 

72. (a + 6)/c + d is equivalent to which of the 
following mathematical expressions? 

a + b 



(a) a H h d 

c 

, . a + 6 

(c) + d 



(b) 



c + d 



(d)o- 



c + d 



73. a + b/(c + d) is equivalent to which of the 
following mathematical expressions? 

, + b 



(a) a H h d 

c 

(c) + d 



(b) 

c + d 

(d)o- 



c c + d 

74. (a + 6)/(c + d) is equivalent to which of 
the following mathematical expressions? 

a + b 



(a) a H h d 

c 

, . a + 6 
(c) + d 



(d) a + - 



75. Use the graphing calculator to reduce 
4125/1155 to lowest terms. 



76. Use the graphing calculator to reduce 
2100/945 to lowest terms. 



40 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



77. Use the graphing calculator to simplify 

45 70 
84 ' 33' 

78. Use the graphing calculator to simplify 

34 13 
55 + 77' 



79. Use the graphing calculator to simplify 



28 
33 



35 

44 



80. Use the graphing calculator to simplify 



11 

84 



11 
30 



f». f». 5». 



Answers 



■*i ■*; -m 



2 
1. - 

5 

3. A 

24 



15 
9 

To 



4 
9 -3 

11. 3 - 

2 



13. 



15. 



45 
13 

171 

"20" 



10 
19. — 

49 



21. 



23. 



120 

'275 

270 
"29" 



25. 


580 
~"39" 


27. 


930 

~289 


29. 


98 
05 


31. 


7 
~12 


33. 


11 

~~9 


35. 


1 

~30 


37. 


70 
~45 


39. 


109 
~"90" 


41. 


79 
30 


43. 


53 
35 



45. 



47. 



131 

"1T 

323 

"^0" 



02 
49. — 

45 



1.3. THE RATIONAL NUMBERS 47 



5 16 

51. 65. 

42 15 

53 _73 67. -64/9 
12 

43 

37 69. — 
55.-— 3 

16 



57. «! 

16 



71. (a) 

73. (d) 

2 
59 - "g 75. 25/7 

43 77. 25/22 

ol. 

10 

79. 16/15 
63. -1/8 



48 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Change 24/7 to a decimal. 



1.4 Decimal Notation 

Every rational number can be expressed using decimal notation. To change a 
fraction into its decimal equivalent, divide the numerator of the fraction by its 
denominator. In some cases the process will terminate, leaving a zero remain- 
der. However, in other cases, the remainders will begin to repeat, providing a 
decimal representation that repeats itself in blocks. 



EXAMPLE 1. 

, , 39 



Change each of the following fractions into decimals. 

v ; 11 



Solution: We perform two divisions, the one on the left to change 39/80 to a 
decimal, the one on the right to find a decimal representation for 4/11. 

0.3636 



0.4875 


80)39.0000 


32 


7 00 


6 40 


600 


560 


400 


400 



1)4.0000 


33 


70 


66 


40 


33 


70 


66 



4 

On the left, the division process terminates with a zero remainder. Hence, 
39/80 = 0.4875 is called a terminating decimal. On the right, the remainders 
repeat in a pattern and the quotient also repeats in blocks of two. Hence, 
4/11 = 0.3636 ... is called a repeating decimal. We can also use a repeating 
bar to write 4/11 = 0.36. The block under the repeating bar repeats itself 
Answer: 3.428571 indefinitely. 

□ 



You Try It! 



Change 0.45 to a fraction. 
Reduce to lowest terms. 



Vice-versa, any terminating decimal can be expressed as a fraction. You 
need only count the number of digits after the decimal point and use the same 
number of zeros in your denominator. 



EXAMPLE 2. Express each of the following decimals as fractions. Reduce 
your answers to lowest terms. 



(a) 0.055 



(b) 3.36 



Solution: In each case, count the number of digits after the decimal point and 
include an equal number of zeros in the denominator. 



1.4. DECIMAL NOTATION 49 

In example (a), there are three dig- In example (b), there are two digits 

its after the decimal point, so we after the decimal point, so we place 

place the number over 1000, which the number over 100, which has two 

has three zeros after the one. zeros after the one. 

55 336 

0.055 = 3.36 = 

1000 100 

11 _ 84 

~ 200 ~ 25 



Answer: 9/20 
□ 



As we saw in Example 1, the repeating decimal 0.36 is equivalent to the 
fraction 4/11. Indeed, any repeating decimal can be written as a fraction. For 
example, 0.3 = 1/3 and 0.142857 = 1/7. In future courses you will learn a 
technique for changing any repeating decimal into an equivalent fraction. 

However, not all decimals terminate or repeat. For example, consider the 
decimal 

0.42422422242222..., 

which neither terminates nor repeats. This number cannot be expressed using 
repeating bar notation because each iteration generates one additional 2. Be- 
cause this number neither repeats nor terminates, it cannot be expressed as a 
fraction. Hence, 0.42422422242222... is an example of an irrational number. 

Irrational numbers. If a number cannot be expressed in the form p/q, where 
p and q are integers, q ^ 0, then the number is called an irrational number. 



Real numbers. By including all of the rational and irrational numbers in one 
set, we form what is known as the set of real numbers. 

The set of real numbers includes every single number we will use in this text- 
book and course. 

Adding and Subtracting Decimals 

When adding signed decimals, use the same rules you learned to use when 
adding signed integers or fractions. 



Sign rules for addition. 


When adding two decimal numbers, 


use the follow- 


ing rules: 










• To add two decimals with like 


signs, add their magnitudes and 


prefix 


their common sign. 











50 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



• To add two decimals with unlike signs, subtract the smaller magnitude 
from the larger, then prefix the sign of the decimal number having the 
larger magnitude. 



You Try It! 



Simplify: -22.6+18.47 



Answer: 



-4.13 



EXAMPLE 3. Simplify: (a) -2.3 + (-0.015) and (b) 



6.95 



Solution: In part (a), note that we have like signs. Hence, we add the mag- 
nitudes and prefix the common sign. 

2.300 

+0.015 
-2.3 + (-0.015) = -2.315 2315~ 

In part (b), note that we have unlike signs. Thus, we first subtract the smaller 
magnitude from the larger magnitude, then prefix the sign of the decimal num- 
ber with the larger magnitude. 

8.40 
-6.95 



6.95 



-1.45 



1.45 



Hence, -2.3 + (-0.015) = -2.315 and -8.4 + 6.95 = -1.45. 



□ 



You Try It! 



Simplify: -22.6 - 18.47 



Answer: 



-41.07 



Subtraction still means "add the opposite." 



EXAMPLE 4. Simplify: (a) -5.6 - 8.4 and (b) -7.9 - (-5.32) 

Solution: In part (a), first we add the oppposite. Then we note that we have 
like signs. Hence, we add the magnitudes and prefix the common sign. 

5.6 

-5.6-8.4 = -5.6 +(-8.4) +8.4 

= -14.0 14-0 

In part (b), first we add the opposite. Then we note that we have unlike signs. 
Thus, we first subtract the smaller magnitude from the larger magnitude, then 
prefix the sign of the decimal number with the larger magnitude. 

7.90 
-7.9- (-5.32) = -7.9 + 5.32 -5.32 

= -2.58 2.58 



Hence, —5.6 



-14.0 and -7.9- (-5.32) 



-2.58. 



□ 



1.4. DECIMAL NOTATION 



51 



Multiplication and Division of Decimals 

The sign rules for decimal multiplication and division are the same as the sign 
rules used for integers and fractions. 



Sign Rules for multiplication and division. When multiplying or dividing 
two decimal numbers, use the following rules: 

• Like signs give a positive result. 

• Unlike signs give a negative result. 



Multiplication of decimal numbers is fairly straightforward. First multiply 
the magnitudes of the numbers, ignoring the decimal points, then count the 
number of digits to the right of the decimal point in each factor. Place the 
decimal point in the product so that the number of digits to the right of the 
decimal points equals the sum of number of digits to the right of the decimal 
point in each factor. 



EXAMPLE 5. Simplify: (-1.96)(2.8) 

Solution: Multiply the magnitudes. The first decimal number has two digits 
to the right of the decimal point, the second has one digit to the right of the 
decimal point. Thus, we must place a total of three digits to the right of the 
decimal point in the product. 

1.96 
x2.8 
1 568 
3 92 

5.488 



(-1.96)(2.8) = -5.488 



Note that unlike signs yield a negative product. 



You Try It! 



Simplify: (-12.5)(-23.4) 



Answer: 292.50 



When dividing signed decimal numbers, ignore the signs and divide the 
magnitudes. Push the decimal point in the divisor to the end of the divisor, 
then move the decimal point in the dividend an equal number of spaces. This 
sets the decimal point in the quotient. 



□ 



You Try It! 



EXAMPLE 6. Simplify: -4.392 -r (-0.36) 

Solution. Divide the magnitudes. Move the decimal in the divisor to the end 
of the divisor. Move the decimal in the dividend an equal number of places 
(two places) to the right. 



Simplify: -5.76/3.2 



52 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



0.36 4.39 2 



Answer: 



Place the decimal point in the quotient directly above the new position of the 
decimal point in the dividend, then divide. 

12.2 



36)439.2 
36 
79 


72 
72 


72 



Like signs yield a positive result. Hence, —4.392-=- (—0.36) = 12.2. 



□ 



Order of Operations 

Decimal numbers obey the same Rules Guiding Order of Operations as do the 
integers and fractions. 



Rules Guiding Order of Operations. When evaluating expressions, pro- 
ceed in the following order. 

1. Evaluate expressions contained in grouping symbols first. If grouping 
symbols are nested, evaluate the expression in the innermost pair of 
grouping symbols first. 

2. Evaluate all exponents that appear in the expression. 

3. Perform all multiplications and divisions in the order that they appear 
in the expression, moving left to right. 

4. Perform all additions and subtractions in the order that they appear in 
the expression, moving left to right. 



You Try It! 



Given y = —0.2, evaluate: 



EXAMPLE 7. Given x = -0.12, evaluate -x 2 . 

Solution: Following Tips for Evaluating Algebraic Expressions, first replace all 
occurrences of variable x in the expression — x 2 with open parentheses. Next, 



1.4. DECIMAL NOTATION 



53 



substitute —0.12 for x in the open parentheses, then simplify. 

\2 



-X 



-(-0.12) z 

-(0.0144) 

-0.0144 



Replace x with open parentheses. 
Substitute —0.12 for x. 
Exponent: (-0.12) 2 = 0.0144 
Negate. 



Note that we square first, then we negate second. Thus, if x = —0.12, then 
-x 2 = -0.0144. 



Answer: 



-0.0016 



□ 



EXAMPLE 8. Given 



-0.3, evaluate 1.2a: 2 — 3 Ax. 



Solution: Following Tips for Evaluating Algebraic Expressions, first replace all 
occurrences of variable x in the expression 1.2a; 2 — 3.4a; with open parentheses. 
Next, substitute —0.3 for x in the open parentheses, then simplify. 



You Try It! 



Given y = —0.15, evaluate: 
-lAy 2 + 2.2y 



1.2a; 2 - 3.4a; 



1.2( ) 2 -3.4( ) 
1.2(-0.3) 2 -3.4(-0.3) 
1.2(0.09) -3.4(-0.3) 
0.108- (-1.02) 

0.108+1.02 
1.128 



Replace x with parentheses. 

Substitute —0.3 for x. 

Exponent: (-0.3) 2 = 0.09. 

Multiply: 1.2(0.09) = 0.108 and 
3.4(-0.3) = -1.02. 

Add the opposite. 

Simplify. 



Thus, if x = -0.3, then 1.2a; 2 - 3.4a; = 1.128. 



Answer: 



-0.3615 



□ 



We saw earlier that we can change a fraction to a decimal by dividing. 



You Try It! 



EXAMPLE 9. Given x = 2/5, evaluate -3.2.x + 5. Given y = -3/4, evaluate: 

Solution: Following Tips for Evaluating Algebraic Expressions, first replace _« o . 7 

all occurrences of variable x in the expression —3.2a; + 5 with open parentheses. 
Next, substitute 2/5 for x in the open parentheses. 



-3.2a; + 5 = -3.2 



Replace x with open parentheses. 



-3.2 ( - ) + 5 Substitute 2/5 for x. 



54 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: 8.725 



You Try It! 



Given z = —0.4, evaluate: 



4 
5- -z 
5 



One approach is to change 2/5 to a decimal by dividing the numerator by the 
denominator. Thus, 2/5 = 0.4. 

= -3.2(0.4) + 5 Replace 2/5 with 0.4. 

= -1.28 + 5 Multiply: -3.2(0.4) = -1.28. 

= 3.72 Add: -1.28+5 = 3.72. 

Thus, if x = 2/5, then -3.2a; + 5 = 3.72. 

□ 

As we saw in Example 2, we can easily change a terminating decimal into 
a fraction by placing the number (without the decimal point) over the proper 
power of ten. The choice of the power of ten should match the number of digits 
to the right of the decimal point. For example: 

411 , _ 311 151111 

100 



0.411 



and 



3.11 



and 



15.1111 



1000 100 10000 

Note that the number of zeros in each denominator matches the number of 
digits to the right of the decimal point. 



EXAMPLE 10. Given y 



-0.25, evaluate y + 4. 



Solution: Following Tips for Evaluating Algebraic Expressions, first replace all 
occurrences of variable y in the expression — (3/5)y + 4 with open parentheses. 
Next, substitute —0.25 for y in the open parentheses. 



:2/ + 4 



(-0.25) 



Replace y with open parentheses. 
Substitute —0.25 for y. 



Answer: 133/25 



Place 25 over 100 to determine that —0.25 
-0.25 = -1/4. 



-25/100, or after reduction, 



1 



20 

3 80 

20 + 20 
83 

20 



+ 4 Replace -0.25 with -1/4. 

Multiply: -| (-1) = A. 

Make equivalent fractions with LCD. 
Add. 



Thus, if y = -0.25, then -(3/5)j/ + 4 = 83/20. 



□ 



1.4. DECIMAL NOTATION 



55 



Rounding Using the Graphing Calculator 

Here is the algorithm for rounding a decimal number to a particular place. 

Rules for rounding. To round a number to a particular place, follow these 
steps: 

1. Mark the place you wish to round to. The digit in this place is called the 
rounding digit. 

2. Mark the digit in the place to the immediate right of the rounding digit. 
This is called the test digit. 

a) If the test digit is greater than or equal to 5, add 1 to the rounding 
digit, then replace all digits to the right of the rounding digit with 
zeros. Trailing zeros to the right of the decimal point may be deleted. 

b) If the test digit is less than 5, keep the rounding digit the same, 
then replace all digits to the right of the rounding digit with zeros. 
Trailing zeros to the right of the decimal point may be deleted. 



EXAMPLE 11. Use your graphing calculator to evaluate 125a; 3 — 17.5a;+44. 
at x = —3.13. Round your answer to the nearest tenth. 



Solution. First, store —3.13 in the variable X with the following keystrokes. 



0QQ0 



X,T,e,n I ENTER 



The result is shown in the first image in Figure 1.15. Next, enter the expression 
125a; 3 — 17. 5x + 44.8 with the following keystrokes. 



ED LUSH 



X,T,0,n 






ENTER 



The result is shown in the second image in Figure 1.15. 

Thus, the answer is approximately —3733.462125. We now need to round 
this answer to the nearest tenth. Mark the rounding digit in the tenths place 
and the test digit to its immediate right. 



Test digit 



Rounding digit 



3733. |T| [t\ 2125 

A 



You Try It! 



Evaluate a; 3 — 3a; at 

x = -1.012. Round to the 

nearest hundredth. 



56 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 




125*X A 3-17.5*X+4 
4.S 

-3733.462125 



Figure 1.15: Evaluate 125a; 3 - 17.5a; + 44.8 at x = -3.13. 



Because the test digit is greater than or equal to 5, add 1 to the rounding digit, 
then replace all digits to the right of the rounding digit with zeros. 



-3733.462125: 



-3733.500000 



Answer: 2.0 



Delete the trailing zeros from end of the fractional part of a decimal. This does 
not change our answer's value. 

-3733.462125 « -3733.5 
Therefore, if x = —3.13. then to the nearest tenth: 



125a; 3 - 17.5a; + 44.6 



-3733.5 



□ 



1.4. DECIMAL NOTATION 



57 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



In Exercises 1-33, simplify the given expression. 

1. -2.835 +(-8.759) 

2. -5.2 +(-2) 

3. 19.5- (-1.6) 

4. 9.174- (-7.7) 

5. -2-0.49 

6. -50.86-9 

7. (-1.2)(-0.05) 

8. (-7.9)(0.9) 

9. -0.13 + 23.49 

10. -30.82 + 75.93 

11. 16.4 +(-41.205) 

12. -7.8 + 3.5 

13. -0.4508 + 0.49 

14. 0.2378 +(-0.29) 

15. (-1.42)(-3.6) 

16. (-8.64) (4.6) 



17. 2.184+ (-0.24) 

18. 7.395 + (-0.87) 

19. (-7.1)(-4.9) 

20. (5.8)(-1.9) 

21. 7.41 + (-9.5) 

22. -1.911 + 4.9 

23. -24.08 + 2.8 

24. 61.42 +(-8.3) 

25. (-4.04)(-0.6) 

26. (-5.43)(0.09) 

27. -7.2- (-7) 

28. -2.761- (-1.5) 

29. (46.9)(-0.1) 

30. (-98.9) (-0.01) 

31. (86.6)(-1.9) 

32. (-20.5)(8.1) 



In Exercises 33-60, simplify the given expression. 

33. -4.3- (-6.1)(-2.74) 

34. -1.4- 1.9(3.36) 

35. -3.49+ |-6.9- (— 15.7)| 

36. 1.3 + | - 13.22-8.79| 

37. |18.9 - 1.55| - | - 16.1 - (-17.04) | 

38. | - 17.5 - 16.4| - | - 15.58 - (-4.5) | 

39. 8.2- (-3.1) 3 

40. -8.4- (-6.8) 3 



41. 5.7- (-8.6)(1.1) 2 

42. 4.8-6.3(6.4) 2 

43. (5.67)(6.8)- (1.8) 2 

44. (-8.7)(8.3)- (-1.7) 2 

45. 9.6 +(-10.05- 13.16) 

46. -4.2 + (17.1- 14.46) 

47. 8.1 + 3.7(5.77) 

48. 8.1 + 2.3(-5.53) 



58 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



49. 7.5 + 34.5/(-1.6 + 8.5) 

50. -8.8 + 0.3/(-7.2 + 7.3) 

51. (8.0 + 2.2)/5.1 - 4.6 

52. (35.3 + 1.8)/5.3- 5.4 

53. -18.24- | - 18.5- 19. 7| 

54. 16.8- 14.58- 17.141 



55. -4.37- | -8.97| 

56. 4.1- | -8.4| 

57. 7.06- (-1.1-4.41) 

58. 7.74- (0.9- 7.37) 

59. -2.2- (-4.5) 2 

60. -2.8- (-4.3) 2 



61. Evaluate a — b 2 at a = —2.9 and b = —5.4. 

62. Evaluate a — b 3 at a = —8.3 and b = —6.9. 

63. Evaluate a+\b-c\ at a = -19.55, b = 5.62, 
and c = —5.21. 



64. Evaluate a — \b — c\ at a 
-8.31, and c = 17.5. 



.37, 



65. Evaluate a — be at a = 4.3, b = 8.5, and 
c= 1.73. 

66. Evaluate a + be at a = 4.1, b = 3.1, and 
c= -7.03. 

67. Evaluate a — (b — c) at a = —7.36, b = 
-17.6, and c= -19.07. 



68. Evaluate \a — b\ — \c — d\ at a = 1.91, 
b = 19.41, c = -11.13, and d = 4.3. 

69. Evaluate a + b/(c + d) at a = 4.7, b = 54.4, 
c = 1.7, and d = 5.1. 

70. Evaluate (a + b)/c - d at a = -74.2, 
6 = 3.8, c = 8.8, andd= 7.5. 

71. Evaluate ab — c 2 at a = —2.45, b = 5.6, 
and c = —3.2. 

72. Evaluate a + (b - c) at a = 12.6, b = 
-13.42, andc= -15.09. 

73. Evaluate a- \b\ at a = -4.9 and b = -2.67. 

74. Evaluate a - be 2 at a = -3.32, b = -5.4, 
and c = —8.5. 



75. Use your graphing calculator to evaluate 3.5 — 1.7x at x = 1.25 Round your answer to the nearest 
tenth. 

76. Use your graphing calculator to evaluate 2.35a: — 1.7 at x = —12.23 Round your answer to the 
nearest tenth. 

77. Use your graphing calculator to evaluate 1.7a; — 3.2x + 4.5 at x = 2.86 Round your answer to the 
nearest hundredth. 

78. Use your graphing calculator to evaluate 19.5 — 4.4a; — 1.2a; 2 at x = —1.23 Round your answer to 
the nearest hundredth. 

79. Use your graphing calculator to evaluate —18.6 + 4.4a: 2 — 3.2a; 3 at x = 1.27 Round your answer 
to the nearest thousandth. 

80. Use your graphing calculator to evaluate —4.4a; 3 — 7.2a; — 18.2 at x = 2.29 Round your answer to 
the nearest thousandth. 



1.4. DECIMAL NOTATION 59 

s*- j*- £*■ Answers <#$ >#s <#$ 

1. -11.594 41. 16.106 

3. 21.1 43. 35.316 

5. -2.49 45. -13.61 

7. 0.06 47. 29.449 

9. 23.36 49. 12.5 

11. -24.805 51. -2.6 

13. -0.92 53. -56.44 

15. 5.112 55. -13.34 

17. -9.1 57. 12.57 

19. 34.79 59. -22.45 

21. -0.78 61. -32.06 

23. -8.6 63. -8.72 

25. 2.424 65. -10.405 

27. -0.2 67. -8.83 

29. -4.69 69. 12.7 

31. -164.54 71. -23.96 

33. -21.014 73. -7.57 

35. 5.31 75. 1.4 

37. 16.41 77. 9.25 

39. 37.991 79. -4.948 



c,0 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: 2(3x) 



Answer: 6x 



1.5 Algebraic Expressions 

The associative property of multiplication is valid for all numbers. 

Associative Property of Multiplication. Let a, b, and c be any numbers. 
Then: 

a ■ (b ■ c) = (a ■ b) ■ c 

The associative property of multiplication is useful in a number of situations. 



EXAMPLE 1. Simplify: -3 (Ay) 

Solution: Currently the grouping — 3(Ay) demands that we first multiply 
4 and y. However, we can use the associative property of multiplication to 
regroup, first multiplying —3 and 4. 



-3(Ay) = (-3 ■ A)y 
= -12y 

Thus, -3(4y) = -12y. 



The associative property of multiplication. 
Multiply: -3-4= -12 



□ 



Let's look at another example. 



You Try It! 



Simplify: -3(-8u 2 



Answer: 2Au 2 



EXAMPLE 2. Simplify: -2(-Axy) 

Solution: Currently, the grouping — 2(— Axy) demands that we first multiply 
—4 and xy. However, we can use the associative property of multiplication to 
regroup, first multiplying —2 and —4. 

— 2(— Axy) = (—2 • (—A))xy The associative property of multiplication. 
= %xy Multiply: -2 ■ (-4) = 8 

Thus, — 2(— Axy) = 8xy. 
□ 

In practice, we can move quicker if we perform the regrouping mentally, then 
simply write down the answer. For example: 



-2(-4t) = 8t and 2(-5z 2 ) = -10z 2 and - 3(4w 3 



I2u j 



1.5. ALGEBRAIC EXPRESSIONS 



61 



The Distributive Property 

We now discuss a property that couples addition and multiplication. Consider 
the expression 2 • (3 + 5). The Rules Guiding Order of Operations require that 
we first simplify the expression inside the parentheses. 



2 ■ (3 + 5) = 2 • 8 
= 16 



Add: 3 + 5 = 8 
Multiply: 2 • 8 



16 



Alternatively, we can instead distribute the 2 times each term in the parenthe- 
ses. That is, we will first multiply the 3 by 2, then multiply the 5 by 2. Then 
we add the results. 



2- (3 + 5) = 2-3 + 2-5 
= 6 + 10 
= 16 



Distribute the 2. 

Multiply: 2 • 3 = 6 and 2 • 5 = 10 

Add: 6+10 = 16 



Note that both methods produce the same result, namely 16. This example 
demonstrates an extremely important property of numbers called the distribu- 
tive property. 



The Distributive Property. Let a, 


b, 


and c be 


any numbers. 


Then: 






a- (b + c) = 


a 


b + a ■ c 






That 


is, multiplication is 


distributive with 


respect 


to addition. 





EXAMPLE 3. Use the distributive property to expand 2(3x + 7). 

Solution: First distribute the 2 times each term in the parentheses. Then 
simplify. 

2(3x + 7) = 2(3x) + 2(7) Use the distributive property. 

= 6a; + 14 Multiply: 2 (3a;) = 6x and 2(7) = 14 

Thus, 2(3x + 7) = 6x + 14. 



You Try It! 



Expand: 5(2y+ 7) 



Answer: Wy + 35 



□ 



Multiplication is also distributive with respect to subtraction. 



You Try It! 



EXAMPLE 4. Use the distributive property to expand — 2(5?/ — 6). 



Expand: -3(2z-7) 



<i2 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Solution: Change to addition by adding the opposite, then apply the dis- 
tributive property. 

-2(5y - 6) = -2(5y + (-6)) Add the opposite. 

= — 2(5y) + (— 2)(— 6) Use the distributive property. 

= -Wy+ 12 Multiply: -2(5y) = -Wy 

and (-2)(-6) = 12 

Answer: -6z + 21 Thus, -2(5y - 6) = -Wy + 12. 

□ 



Some might prefer to "take 
their time," using the 
technique shown in 
Examples 3 and 4 until they 
feel "ready" to move more 
quickly. 



Speeding Things Up a Bit 

In Example 4, we changed the subtraction to addition, applied the distributive 
property, then several steps later we were finished. However, if you understand 
that subtraction is really the same as adding the opposite, and if you are willing 
to do a few steps in your head, you should be able to simply write down the 
answer immediately following the given problem. 

If you look at the expression — 2(5y — 6) from Example 4 again, only this 
time think "multiply —2 times by, then multiply —2 times —6, then the result 
is immediate. 

-2(5y-6) = -10y+12 

Let's try this "speeding it up" technique in a couple more examples. 



You Try It! 



-2x - 



12) 



Expand: — 3(— 2a + 36—7) EXAMPLE 5. Use the distributive property to expand — 3(- 

Solution: To distribute the —3, we simply think as follows: "— 3(— 2x) = 6x, 
— 3(5y) = — 15y, and — 3(— 12) = 36." This sort of thinking allows us to write 
down the answer immediately without any additional steps. 



-5y- 

-2x) 



Answer: 6a — 96 + 21 



-3(-2x + 5y- 12) = 6a; - 15y + 36 



□ 



You Try It! 



Expand: -4(— x - 2y - 7) 



Answer: 4a; + 8y + 28 



EXAMPLE 6. Use the distributive property to expand — 5(— 2a — 56 + 8). 

Solution: To distribute the —5, we simply think as follows: "— 5(— 2a) = 10a, 
— 5(— 56) = 256, and —5(8) = —40." This sort of thinking allows us to write 
down the answer immediately without any additional steps. 



-5(-2a-56- 



10a + 256 -40 



□ 



1.5. ALGEBRAIC EXPRESSIONS 63 

Distributing a Negative Sign 

Recall that negating a number is equivalent to multiplying the number by — 1 . 

Multiplicative Property of Minus One. If a is any number, then: 

(— l)a = —a 



This means that if we negate an expression, it is equivalent to multiplying the 
expression by —1. 



You Try It! 



EXAMPLE 7. Expand -{7x- 8y - 10). Expand: -(-a -26 +11) 

Solution: First, negating is equivalent to multiplying by —1. Then we can 
change subtraction to addition by "adding the opposite" and use the distribu- 
tive property to finish the expansion. 

— (7x — Sy — 10) = — l(7a; — 8y — 10) Negating is equivalent to 

multiplying by — 1 

= -l(7x + (-8y) + (-10)) Add the opposite. 

= -l(7x) + (-l)(-8y) + (-1)(-10) Distribute the -1. 

= -7x + 8y + 10 Multiply. 

Thus, -(7x -8y- 10) = -7x + 8y + 10. Answer: a + 26 - 11 
□ 



While being mathematically precise, the technique of Example 7 can be simpli- 
fied by noting that negating an expression surrounded by parentheses simply 
changes the sign of each term inside the parentheses to the opposite sign. 



Once we understand this, we can simply "distribute the minus sign" and write: 

-(7a; -8y- 10) = -7x + 8y + 10 

In similar fashion, 

— (—3a + 56 — c) = 3a — 56 + c 

and 

-(-3a; - 8y + 11) = 3x + 8y- 11. 



04 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: 3y + 8y 



Answer: lly 



Combining Like Terms 

We can use the distributive property to distribute a number times a sum. 

a(b + c) = ab + ac 

However, the distributive property can also be used in reverse, to "unmultiply" 
or factor an expression. Thus, we can start with the expression ab + ac and 
"factor out" the common factor a as follows: 

ab + ac = a(b + c) 
You can also factor out the common factor on the right. 

ac + be = (a + b)c 
We can use this latter technique to combine like terms. 

EXAMPLE 8. Simplify: 7x + 5x 

Solution: Use the distributive property to factor out the common factor x 
from each term, then simplify the result. 



7a; + 5a; = (7 + 5)a; 

= 12x 

Thus, 7a; + 5a; = 12a;. 



Factor out an x using 
the distributive property. 

Simplify: 7 + 5 = 12 



□ 



You Try It! 



Simplify: — 5z + 9z 



3 _!_ o~3 



EXAMPLE 9. Simplify: -8a 2 + 5a 2 

Solution: Use the distributive property to factor out the common factor o 2 
from each term, then simplify the result. 



Answer: 4^ 



la 2 + 5a 2 = (-8 + 5)a 2 



-3a 2 



Factor out an a using 
the distributive property. 

Simplify: —8 + 5 = —3 



Thus, -8a 2 + 5a 2 



-3a 2 



a 

Examples 8 and 9 combine what are known as "like terms." Examples 8 
and 9 also suggest a possible shortcut for combining like terms. 



1.5. ALGEBRAIC EXPRESSIONS 



65 



Like Terms. Two terms are called like terms if they have identical variable 
parts, which means that the terms must contain the same variables raised to 
the same exponents. 

For example, 2x 2 y and llx 2 y are like terms because they contain identical 
variables raised to the same exponents. On the other hand, — 3si 2 and As 2 t 
are not like terms. They contain the same variables, but the variables are not 
raised to the same exponents. 

Consider the like terms 2x 2 y and llx 2 y. The numbers 2 and 11 are called 
the coefficients of the like terms. We can use the distributive property to 
combine these like terms as we did in Examples 8 and 9, factoring out the 
common factor x 2 y. 

2x 2 y + llx 2 y= (2 + ll)x 2 y 
= 13x 2 y 

However, a much quicker approach is simply to add the coefficients of the like 
terms, keeping the same variable part. That is, 2 + 11 = 13, so: 



2x 2 y + llx 2 y 



13x 2 y 



This is the procedure we will follow from now on. 



EXAMPLE 10. Simplify: -8w 2 + 17w 2 

Solution: These are like terms. If we add the coefficients —8 and 17, we get 
9. Thus: 



-8w 2 + 17w 2 



9w 2 



Add the coefficients and 
repeat the variable part. 



You Try It! 



Simplify: Aab — \bab 



Answer: — llafc 



a 



EXAMPLE 11. Simplify: -Auv - 9uv 

Solution: These are like terms. If we add —4 and —9, we get —13. Thus: 

— Auv — 9uv = —I3uv Add the coefficients and 

repeat the variable part. 



You Try It! 



Simplify: —3xy — 8xy 



Answer: — llxy 



a 



You Try It! 



EXAMPLE 12. Simplify: -3x 2 y + 2xy 2 



Simplify: 5ab + llbc 



(i(» 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



Answer: 5ab + llbc 



You Try It! 



Simplify: 



-3z 2 + Az- 8z 2 -9z 



Solution: These are not like terms. They do not have the same variable 
parts. They do have the same variables, but the variables are not raised to the 
same exponents. Consequently, this expression is already simplified as much 
as possible. 



-3x 2 y + 2xy 2 



Unlike terms. Already simplified. 



□ 

Sometimes we have more than just a single pair of like terms. In that case, 
we want to group together the like terms and combine them. 



EXAMPLE 13. Simplify: -8u - Av - 12m + 9v 

Solution: Use the associative and commutative property of addition to change 
the order and regroup, then combine line terms. 

— 8m — Av — 12m + 9v = (—8m — 12m) + (—Av + 9v) Reorder and regroup. 
= — 20m + 5m Combine like terms. 



Answer: — 11 z 2 — 5z 



Note that -8m- 12m 



-20m and — Av + 9v = 5m. 



Alternate solution. You may skip the reordering and regrouping step if you 
wish, simply combining like terms mentally. That is, it is entirely possible to 
order your work as follows: 



-8m- Av- 12m + 9m 



-20m + 5m 



Combine like terms. 



□ 

In Example 13, the "Alternate solution" allows us to move more quickly 
and will be the technique we follow from here on, grouping and combining 
terms mentally. 



Order of Operations 

Now that we know how to combine like terms, let's tackle some more compli- 
cated expressions that require the Rules Guiding Order of Operations. 



Rules Guiding Order of Operations. When evaluatin 


g expressions, pro- 


ceed 


in the following order. 






1. 


Evaluate expressions contained in 


grouping symbols 


first. If grouping 




symbols are nested, evaluate the 


expression in the 


innermost pair of 




grouping symbols first. 







1.5. ALGEBRAIC EXPRESSIONS 



67 



2. Evaluate all exponents that appear in the expression. 

3. Perform all multiplications and divisions in the order that they appear 
in the expression, moving left to right. 

4. Perform all additions and subtractions in the order that they appear in 
the expression, moving left to right. 



EXAMPLE 14. Simplify: 4(-3a + 26) - 3(4a - 56) 

Solution: Use the distributive property to distribute the 4 and the —3, then 
combine like terms. 

4(-3a + 26) - 3(4a - 56) = - 12a + 86 - 12a + 156 Distribute. 

= —24a + 236 Combine like terms. 



Note that -12a- 12a: 



-24a and 



156 = 236. 



You Try It! 



Simplify: 



-2x - 3(5 - 2cc) 



Answer: 4a; — 15 



□ 



You Try It! 



EXAMPLE 15. Simplify: -2(3x - Ay) - (5a; - 2y) 

Solution: Use the distributive property to multiply —2 times 3a; — Ay, then 
distribute the minus sign times each term of the expression 5a;— 2y. After that, 
combine like terms. 

-2(3cc - Ay) - (5a: - 2y) = -6x + 8y- 5x + 2y Distribute. 

= — 11a; + 10y Combine like terms: 

Note that —6a; — 5a; = —lis and 8y + 2y = lOy. 



Simplify: 



-3(u + v) — (u — 5v) 



Answer: — Au + 2v 



□ 



EXAMPLE 16. Simplify: -2{x 2 y - 3a;y 2 ) - A(-x 2 y + ixy 2 ) 

Solution: Use the distributive property to multiply —2 times x 2 y — 3xy 2 and 
—4 times —x 2 y + 3a;y 2 . After that, combine like terms. 

-2{x 2 y - 3xy 2 ) - A(-x 2 y + 3xy 2 ) = -2x 2 y + 6xy 2 + Ax 2 y - 12xy 2 

= 2x y — 6xy 

Note that —2x 2 y + Ax 2 y = 2x 2 y and 6a;y 2 — 12a;y 2 = —6xy 2 . 



You Try It! 



Simplify: 



8u 2 v - 3(u 2 v + Auv 2 ) 



Answer: 5u 2 v — 12uv 2 



D 



G8 



CHAPTER 1. THE ARITHMETIC OF NUMBERS 



You Try It! 



Simplify: 

x - 2[-x + 4(x + 1)} 



When grouping symbols are nested, evaluate the expression inside the in- 
nermost pair of grouping symbols first. 



EXAMPLE 17. Simplify: -2x - 2 (-2x - 2 [-2x - 2]) 

Solution: Inside the parentheses, we have the expression — 2x — 2[— 2a; — 2]. 
The Rules Guiding Order of Operations dictate that we should multiply first, 
expanding —2 [—2x — 2] and combining like terms. 

-2x - 2 (-2.T - 2 [-2x - 2]) = -2x - 2 (-2x + Ax + 2) 

= -2x- 2 (2a; + 2) 

In the remaining expression, we again multiply first, expanding — 2(2a; + 2) and 
combining like terms. 



Answer: — 5x 



-2a; - 4a; - 4 
-6a; -4 



□ 



1.5. ALGEBRAIC EXPRESSIONS 69 



**.**.**. Exercises ■*$*$-*$ 

In Exercises 1-6, use the associative property of multiplication to simplify the expression. Note: You 
must show the regrouping step using the associative property on your homewwork. 

1. -3(6o) 4. 8{5xy) 

2. -W(2y) 5. -7(3a; 2 ) 

3. -9(6o6) 6. -6(8z) 



In Exercises 7-18, use the distributive property to expand the given expression. 

7.1(3x-7y) 13. -(-3u - 6v + 8) 

8. -4(5a + 26) 14. -(3u - 3v - 9) 

9. -6(-y + 9) 15. -8(4u 2 - 6v 2 ) 

10. 5(-9u> + 6) 16. -5 (8a; - 9y) 

11. -9(s + 9) 17. -(7u + 10^ + 8) 

12. 6(-10y + 3) 18. -(7u-8v-5) 

In Exercises 19-26, combine like terms by first using the distributive property to factor out the common 
variable part, and then simplifying. Note: You must show the factoring step on your homework. 

19. -19a; + 17a; - 17a; 23. 9y 2 x + Uy 2 x - 3y 2 x 

20. lira- 3n- 18n 24. 4a; 3 - 8a; 3 + 16a; 3 

21. 14a; 3 - 10a; 3 25. 15rra + Um 

22. -lly 3 -6y 3 26. 19 q + 5q 

In Exercises 27-38, simplify each of the following expressions by rearranging and combining like terms 
mentally. Note: This means write down the problem, then write down the answer. No work. 

27. 9- 17m- m + 7 31. -5m - 16 + 5 - 20m 

28. -11 + 20a; + 16a; - 14 32. -18q + 12 - 8 - 19 q 

29. -6y 2 - 3a; 3 + Ay 2 + 3a; 3 33. -16x 2 y + 7y 3 - 12y 3 - 12x 2 y 

30. Uy 3 - lly 2 x + lly 3 + 10y 2 x 34. 10a: 3 + Ay 3 - 13y 3 - 14a; 3 



70 CHAPTER 1. THE ARITHMETIC OF NUMBERS 

35. -Ur + 16 - 7r - 17 37. 14 - 16w - 10 - 13m 

36. -9s- 5- 10s + 15 38. 18 + lOcc + 3 - 18a; 

In Exercises 39-58, use the distributive property to expand the expression, then combine like terms 
mentally. 

39. 3 - (-5m + 1) 49. -7- (5 + 3x) 

40. 5- (-10? + 3) 50. 10 -(6 -4m) 

41. -(9y 2 + 2x 2 ) - 8(5y 2 - 6a; 2 ) 51. -8(-5m - 8) - 7(-2 + Qy) 

42. -8{-8y 2 + Ax 3 ) - 7(3m 2 + x 3 ) 52. 6(-3s + 7) - (4 - 2s) 

43. 2(10 - 6p) + 10(-2p + 5) 53. 4(-7m 2 - 9x 2 y) - 6(-5x 2 y - by 2 ) 

44. 2(3 - 7x) + (-7x + 9) 54. -6(a; 3 + 3y 2 x) + 8(-y 2 x - 9x 3 ) 

45. 4(-10n + 5) - 7(7n - 9) 55. 6s - 7 - (2 - 4s) 

46. 3(-9n + 10) + 6(-7n + 8) 56. 4.x - 9 - (-6 + 5a;) 

47. -Ax - 4 - (10a; - 5) 57. 9(9 - lOr) + (-8 - 2r) 

48. 8y + 9 - (-8y + 8) 58. -7(6 + 2p) + 5(5 - hp) 

In Exercises 59-64, use the distributive property to simplify the given expression. 

59. -7a; + 7(2a; - 5[8ir + 5]) 62. 2x + A(5x - 7[8a; + 9]) 

60. -9a; + 2(5a; + 6 [-8a; - 3]) 63. -8a; - 5(2a; - 3[-4a; + 9]) 

61. 6a; - 4(-3a; + 2[5a; - 7]) 64. 8a; + 6(3a; + 7[-9a; + 5]) 

s*> j*> j*' Answers ■*& •*& •** 

1. -18a 11. -9s -81 

3. -54a6 13. 3u + 6v - 8 

5. -21a; 2 15. -32m 2 + 48w 2 

7. 12a;-28w 17. -7m - 10m - 8 

9. 6m -54 19. -19a; 



1.5. ALGEBRAIC EXPRESSIONS 71 

21. 4a; 3 43. 70 - 32p 

23. 19y 2 x 45. -89n + 83 

25.29m 47. -14a; +1 

27. 16- 18m 49. -12 -3a; 

29. -2y 2 51. -23y + 78 

31. -25m -11 53. 2y 2 - 6x 2 y 

33. -28x 2 y - 5y 3 55. 10s - 9 

35. -21r-l 57. 73 - 92r 

37. 4-29y 59. -273a; - 175 

39. 2 + 5y 61. -22a; + 56 

41. -49y 2 + 46a; 2 63. -78a; + 135 



Chapter 2 

Solving Linear Equations and 
Inequalities 



In this chapter, we will begin solving linear equations by using inverse opera- 
tions to solve for the unknown variable. Humans have been solving problems 
of this type for thousands of years. 

The earliest records indicate that the problems were written entirely in 
words with no symbols (such as +, — , x, -=-, and =) or numbers. For example, 
a problem might read "Sixteen added to an unknown number is fifty-two." 

Diophantes of Alexandria wrote a series of books Arithmetica demonstrat- 
ing solutions to equations in which he used some abbreviations for shortcuts. 

In 9th century AD, al-Khwarizmi wrote the book Al-Kitab al-mukhtasar ti 
Hisab al-jabr w'al-muqabala, translated as The Compendious Book on Calcu- 
lation by Restoration and Reduction. In fact, the word "algebra" is based on 
the word al-Jabr from the text title. His book described the rules of "Comple- 
tion and Reduction" to make it easier to do arithmetic computations needed 
in human business interactions. 

In the 1600-1700's, mathematicians began writing out equations symboli- 
cally as we do today using the symbols +, — , x, and -j- for operations and = 
for equals. 

We will be solving equations which can be solved in logical steps by using 
inverse operations which "undo" each other. For example, one pair of inverse 
operations is addition and subtraction. Another pair of inverse operations is 
multiplication and division. 

We will also be solving inequalities which may have an infinite set of answers 
that can be expressed in three ways: by graphing on a number line, by using 
set-builder notation, or by using interval notation. 



73 



74 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

2.1 Solving Equations: One Step 

Let's start with the definition of an equation. 

Equation. An equation is a mathematical statement that equates two alge- 
braic expressions. 

The key difference between a algebraic expression and an equation is the pres- 
ence of an an equals sign. So, for example, 

2x + 3, x - (3 - 2x), and 2(y + 3) - 3(1 - y) 

are algebraic expressions, while 

2a; + 3 = 0, x - (3 - 2x) = 4, and 2{y + 3) - 3(1 - y) = -11 

are equations. Note that each of the equations contain an equals sign, but the 
algebraic expressions do not. 

Next we have the definition of a solution of an equation. 

What it Means to be a Solution of an Equation. A solution of an 
equation is a numerical value that satisfies the equation. That is, when the 
variable in the equation is replaced by the solution, a true statement results. 



You Try It! 



Which of the numbers 

{1, 2, 3, 4, 5} is a solution of 

the equation 2y + 3 = 7. 



Answer: 



EXAMPLE 1. Show that 8 is a solution of the equation x — 12 = —4. 
Solution. Substitute 8 for x in the given equation and simplify. 



12 = -4 

12 = -4 

-4 = -4 



The given equation. 
Substitute 8 for x. 
Simplify both sides. 



Since the left- and right-hand sides of the last line are equal, this shows that 
when 8 is substituted for x in the equation a true statement results. Therefore, 
8 is a solution of the equation. 

□ 



Equivalent Equations 

Now that we know how to identify a solution of an equation, let's define what 
it means when we say that two equations are equivalent. 



2.1. SOLVING EQUATIONS: ONE STEP 



75 



Equivalent Equations. Two equations are equivalent if they have the same 
solution set. 



You Try It! 



EXAMPLE 2. Are the equations x — 3 = 6 and x = 9 equivalent? 



Solution. The number 9 is the only solution of the equation x — 3 = 6 

Similarly, 9 is the only solution of the equation x = 9. Therefore x — 3 = 6 and 

x = 9 have the same solution sets and are equivalent. Answer 



Are the equations x = 5 and 
X — 7 = 10 equivalent? 



No. 



□ 



You Try It! 



EXAMPLE 3. Are the equations x 2 = 1 and x = 1 equivalent? 
Solution. By inspection, the equation x 2 = 1 has two solutions, —1 and 1. 



Are the equations x = 1 and 
x 2 = x equivalent? 



(-1) 2 



1 



and 



1 



1 



On the other hand, the equation x = 1 has a single solution, namely 1. Hence, 
the equations x 2 = 1 and x = 1 do not have the same solution sets and are 
not equivalent. 



Answer: No. 



As we shall soon see, equivalent equations play an important role in finding 
the solution of an equation. 



□ 



Wrap and Unwrap, Do and Undo 

Suppose that you are wrapping a gift for your cousin. You perform the following 
steps in order. 

1. Put the gift paper on. 

2. Put the tape on. 

3. Put the decorative bow on. 

When we give the wrapped gift to our cousin, he politely unwraps the present, 
"undoing" each of our three steps in inverse order. 

1. Take off the decorative bow. 

2. Take off the tape. 

3. Take off the gift paper. 



76 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

This seemingly frivolous wrapping and unwrapping of a gift contains some 
deeply powerful mathematical ideas. Consider the mathematical expression 
x + 4. To evaluate this expression at a particular value of x, we would start 
with the given value of x, then add 4. 

• Let's set x equal to the number 7. If we add 4, we arrive at the following 
result: 11 

Now, how would we "unwrap" this result to return to our original number? 
We would start with our result, namely 11, then subtract 4. 

• Take our result from above, 11. If we subtract 4, we return to our original 
value of x: 7 

The above discussion leads us to two extremely important observations. 

The inverse of addition is subtraction. If we start with a number x and 
add a number a, then subtracting a from the result will return us to the original 
number x. In symbols, 

x + a — a = x. 

That is, subtracting a "undoes" the effect of adding a and returns us to the 
original number x. 

The inverse of subtraction is addition. If we start with a number x and 
subtract a number a, then adding a to the result will return us to the original 
number x. In symbols, 

x — a + a = x. 

That is, adding a "undoes" the effect of subtracting a and returns us to the 
original number x. 



Operations that Produce Equivalent Equations 

In Example 1, we saw that x = 8 was a solution of the equation x — 12 = —4. 
Indeed, the equations x = 8 and x — 12 = —4 are equivalent equations because 
both have the same solution sets. 

In the case of the equation x — 12 = —4, it's fairly easy to "guess" that 
the solution is x = 8, but as the equations become more complicated, we will 
want to know just what operations we can perform on the equation that will 
not change the solution set. The goal is to start with an equation such as 

a; -12 = -4, 

then through a series of steps that do not change the solution, arrive at the 
solution 



2.1. SOLVING EQUATIONS: ONE STEP 77 

With these thoughts in mind, there are a number of operations that will pro- 
duce equivalent equations (equations with the same solution set). The first 
two that we will employ are adding or subtracting the same amount from both 
sides of an equation. 



Adding the Same Quantity to Both Sides of an Equation. Adding the 
same quantity to both sides of an equation does not change the solution set. 
That is, if 

a = b, 

then adding c to both sides of the equation produces the equivalent equation 

a + c = b + c. 



Subtracting the Same Quantity from Both Sides of an Equation. 

Subtracting the same quantity to both sides of an equation does not change 
the solution set. That is, if 

a = b, 

then subtracting c from both sides of the equation produces the equivalent 
equation 

a — c = b — c. 



Let's look at an example where adding the same amount to both sides of the 
equation produces an equivalent equation that is the solution to the original 
equation. 



You Try It! 



EXAMPLE 4. Solve x - 7 = 12 for x. Solve for 



,r: 



Solution: To undo the effect of subtracting 7, we add 7 to both sides of the x — 6 

equation. 

x — 7 = 12 Original equation. 

x — 7 + 7=12 + 7 Adding 7 to both sides of the equation 
produces an equivalent equation. 

x = 19 On the left, adding 7 "undoes" the effect 

of subtracting 7 and returns x. On the right, 
12 + 7= 19. 

Therefore, the solution of the equation is 19. 



78 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Answer: 10 



You Try It! 



Solve for x: 



1 _ 3 

2 ~~ 5 



Check: To check, substitute the solution 19 into the original equation. 



x-7=12 

19-7 = 12 
12 = 12 



Original equation. 
Substitute 19 for x. 
Simplify both sides. 



The fact that the last line of the check is a true statement guarantees that 19 
is a solution of x — 7 = 12. 

□ 

In the solution of Example 4, we use the concept of the "inverse." If we start 
with x, subtract 7, then add 7, we are returned to the number x. In symbols, 
x — 7 + 7 = x. We are returned to x because "subtracting 7" and "adding 7" 
are inverse operations of one another. That is, whatever one does, the other 
"undoes." 

Next, let's look at an example where subtracting the same amount from 
both sides of the equation produces an equivalent equation that is the solution 
to the original equation. 



EXAMPLE 5. Solve 



for 



Solution: To undo the effect of adding 2/3, we subtract 2/3 from both sides 
of the equation. 



2 1 

x + - = - 

3 2 






_ 3 4 
X ~6~6 



Original equation. 

Subtracting 2/3 from both sides 
produces an equivalent equation. 

On the left, subtracting 2/3 "undoes" 
the effect of adding 2/3 and returns x. 
On the right, make equivalent fractions 
with a common denominator. 

34 ! 
Subtract: = 

6 6 6 



Therefore, the solution of the equation is —1/6. 

Check: Let's use the TI-84 graphing calculator to check this solution. 

1. Store the value —1/6 in the variable X using the following keystrokes. 
The result is shown in Figure 2.1. 



EDQBH 



ENTER 



2.1. SOLVING EQUATIONS: ONE STEP 



79 



2. Enter the left-hand side of the original equation: x + 2/3. Use the fol- 
lowing keystrokes. The result is shown in Figure 2.1. 



X,T,0,n 



B0B0 



-l/6-»X 



1666666667 



Figure 2.1: Checking the solution to x + 2/3 = 1/2. 



3. Press the MATH button on your calculator (see Figure 2.2), then select 
l:^Frac and press the ENTER button. This will convert the decimal 
result to a fraction (see Figure 2.2). 



BJJ1I1 HUN CPtt PRE 


-l/6-»X 


ija^Frac 


-. 1666666667 


2: ►Dec 


X+2/3 


3:* 


.5 


4:*J-< 


flns^Frac 


5: *f 


1/2 


6:fMin<: 


I 


7-i-fMaxC 





Figure 2.2: Changing the result to a fraction. 



1 1 2 1 
Note that the result is — , showing that is a solution of x H — = — . 

2 6 3 2 



Answer: 1/10 



□ 



More Operations That Produce Equivalent Equations 

Here are two more operations that produce equivalent equations. 



Multiplying Both Sides of an Equation by a Nonzero Quantity. Mul- 
tiplying both sides of an equation by a nonzero quantity does not change the 
solution set. That is, if 

a = b, 

and c / 0, then multiplying both sides of the equation by c produces the 
equivalent equation 

ac = be. 



80 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Dividing Both Sides of an Equation by a Nonzero Quantity. Dividing 
both sides of an equation by a nonzero quantity does not change the solution 
set. That is, if 

a = b, 

and c/0, then dividing both sides of the equation by c produces the equivalent 
equation 

a b 



Like addition and subtraction, multiplication and division are inverse opera- 
tions. 



The inverse of multiplication is division. If we start with a number x 
and multiply by a number a, then dividing the result by the same number a 
returns us to the original number x. In symbols, 



That is, dividing by a "undoes" the effect of multiplying by a and returns us 
to the original number x. 

The inverse of division is multiplication. If we start with a number x 
and divide by a number a, then multiplying the result by the same number a 
returns us to the original number x. In symbols, 



That is, multiplying by a "undoes" the effect of dividing by a and returns us 
to the original number x. 



Let's look at an example where dividing both sides of the equation by 
the same amount produces an equivalent equation that is the solution to the 
original equation. 



You Try It! 



Solve for x: 

-3.6a; = 0.072 



EXAMPLE 6. Solve -2.1a; = 0.42 for 



Solution: To undo the effect of multiplying by —2.1, we divide both sides of 



2.1. SOLVING EQUATIONS: ONE STEP 



si 



-2.1a; = 


0.42 


-2.1a; 


0.42 


-2.1 


-2.1 


x = 


-2 



the equation by —2.1. 

Original equation. 

Dividing both sides by —2.1 
produces an equivalent equation. 

On the left, dividing by —2.1 "undoes" the 
effect of multiplying by —2.1 and returns x. 
On the right, divide: 0.42/(-2.1) = -2. 

Therefore, the solution of the equation is —2. 

Check: To check, substitute the solution —2 into the original equation. 

— 2.1a; = 0.42 Original equation. 

-2.1(-2) = 0.42 Substitute -2 for x. 

0.42 = 0.42 On the left, multiply: -2.1(-2) = 0.42 

The fact that the last line of the check is a true statement guarantees that 
is a solution of —2.1a; = 0.42. 



Answer: 



-0.02 



□ 



Next, lets look at an example where multiplying both sides of the equation 
by the same amount produces an equivalent equation that is the solution to 
the original equation. 



EXAMPLE 7. Solve 



-10 for 



Solution: To undo the effect of dividing by 5, we multiply both sides of the 
equation by 5. 



L5 



X 

5 ~~ 


-10 


; J = 


[-10] 5 


x = 


-50 



Original equation. 

Multiplying both sides by 5 
produces an equivalent equation. 

On the left, multiplying by 5 "undoes" the 
effect of dividing by 5 and returns x. On the 
right, multiply: [— 10]5 = —50. 

Therefore, the solution of the equation is —50. 

Check: Let's use the TI-84 graphing calculator to check this solution. 

1. Store the value —50 in the variable X using the following keystrokes. The 
result is shown in Figure 2.3. 



Eomm 



ENTER 



You Try It! 



Solve for x: 



x 

7 



82 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



2. Enter the left-hand side of the original equation: — . Use the following 

5 
keystrokes. 



X,T,0,n 



E 



The result is shown in Figure 2.3. 



-50-»X 
X/5 



-50 
-10 



Figure 2.3: Checking the solution to x/5 = —10. 



Answer: —14 Note that the result is —10, showing that —50 is a solution of x/5 = —10. 



□ 



Writing Mathematics 

When solving equations, observe the following rules when presenting your work. 



1. One 


equation per line. 


This means that 


you s 


hould not 


arrange your 


work 


horizontally as follows: 














x + 3 = 7 


x + 3 


-3 = 7 


-3 


x = 4 






That 


s three equations on 


a line. 


Instead, 


work 


vertically, 


writing 


one 


equat 


ion per line. In the following 


presentation, 


note how 


we align 


the 


equal 


signs in a column. 


















x + 3 = 7 














z + 3- 


3 = 7- 

x = 4 


3 









2.1. SOLVING EQUATIONS: ONE STEP 83 



2. Add and subtract inline. 


Do not add 7 to both sides of the equation 


in the following 


manner: 


x-7 = 
+ 7 
x = 


: 12 

+7 
= 19 




Instead, add 7 ' 


'inline" to both sides of the equation 


as follows: 






x-7 = 


= 12 






X 


-7+7 = 
x = 


= 12 + 7 
= 19 





84 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



;*■;-*■ ti. 



Exercises 



•*;■*; ■*■; 



In Exercises 1-6, which of the numbers following the given equation are solutions of the given equation? 
Support your response with work similar to that shown in Example 1. 



1. x + 2 = 4 

2. £ + 6 = 9 

3. £-9 = 4 



3, 2, 9, 5 

4, 6, 3, 10 
16, 14, 20, 13 



4.x- 3 = 5; 9,15,8,11 

5. a;-3 = 6; 9,16,10,12 

6. x-5 = 7; 15,13,19,12 



In Exercises 7-12, are the following equations equivalent? 



7. £ — 1 = —7 and x = —8 

8. £ — 7 = —8 and x = —15 

9. £ — 5 = —5 and x = 



10. £ — 3 = —3 and x = 

11. £ 2 = 1 and £ = 1 

12. £ 2 = 16 and x = 4 



In Exercises 13-32, solve the given equation for x. 



13. £-20 = 9 

14. £- 10 = 5 

15. 16 = £-3 

16. 16 = £-8 

17. £ + 11 = 20 

18. £ + 10= 18 

19. 9 = 2-19 

20. 4 = £- 11 

21. 20 = 9 + £ 

22. 18 = 7 + £ 



23. 18= 17 + £ 

24. 15 = 7 + £ 

25. 7 + £ = 19 

26. 16 + £ = 17 

27. £ - 9 = 7 

28. £-2 = 8 

29. £ + 15 = 19 

30. £ + 6 = 10 

31. 10 + £ = 15 

32. 18 + £ = 19 



In Exercises 33-40, solve the equation and simplify your answer. 



4 2 
33. £ = - 

9 7 



1 1 
34. £ - - = - 

3 2 



2.1. SOLVING EQUATIONS: ONE STEP 85 



7 4 11 

35.* + - = -- 38 - + I = 3 

19 9 1 

36. x + - = -- 39.x = — 

2 7 8 2 

5 7 5 2 

37. #+- = - 40. a; = — 

9 2 9 3 



In Exercises 41-46, solve the given equation for x. 



41. 


-5.1a; = 


-12.75 


42. 


—3.5a; = 


-22.4 


43. 


-6.9x = 


-58.65 



44. -1.4a; = -4.34 

45. -3.6a; = -24.12 

46. -6.4a; = -39.68 



In Exercises 47-52, solve the given equation for x. 



47. - = -11 
2 

48. - = 12 



49. - = -18 



50. 


1~" 


51. 


^ = * 


52. 


l = -7 



j»- s*/ 5*> Answers >** •** •** 



1. 2 17. 9 

3. 13 19 - 28 



21. 11 
23. 1 

25. 12 
27. 16 
11. No 29. 4 



5. 9 

7. No 
9. Yes 



13. 29 31 - 5 

46 
15. 19 33 - 63 



86 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

43. 8.5 
45. 6.7 
47. -22 
49. -144 
41.2.5 51.-105 



35. 


79 
~36 


37. 


53 
18 


39. 


5 



2.2. SOLVING EQUATIONS: MULTIPLE STEPS 



87 



2.2 Solving Equations: Multiple Steps 

Recall the "Wrap" and "Unwrap" discussion from Section 2.1. To wrap a 
present we put the gift paper on, put the tape on, and put the decorative bow 
on. To unwrap the gift, we must "undo" each of these steps in inverse order. 
Hence, to unwrap the gift we take off the decorative bow, take off the tape, 
and take off the gift paper. 

Now, imagine a machine that multiplies its input by 3, then adds 5 to the 
result. This machine is pictured on the left in Figure 2.4. 



1. Multiply by 3. 

2. Add 5. 



1. Subtract 5. 

2. Divide by 3. 



Figure 2.4: The second machine "unwraps" the first machine. 



To "unwrap" the effect of the machine on the left, we need a machine that 
will "undo" each of the steps of the first machine, but in inverse order. The 
"unwrapping" machine is pictured on the right in Figure 2.4. It will first sub- 
tract 5 from its input, then divide the result by 3. Note that each of these 
operations "undoes" the corresponding operation of the first machine, but in 
inverse order. 

The following argument shows that the second machine "undoes" the op- 
eration of the first machine. 



1. Drop the integer 4 into the machine on the left in Figure 2.4. This ma- 
chine will first multiply 4 by 3, then add 5 to the result. The result is 

3(4) + 5, or 17. 

2. To "unwrap" this result, drop 17 into the machine on the right. This 
machine first subtracts 5, then divides the result by 3. The result is 
(17 — 5)/3, or 4, the original integer that was put into the first machine. 



EXAMPLE 1. Solve for x: 3x + 5 = 14 



You Try It! 



Solve for 



Solution: On the left, order of operations demands that we first multiply x 
by 3, then add 5. To solve this equation for x, we must "undo" each of these 
operations in inverse order. Thus, we will first subtract 5 from both sides of 



2x + 3 = 7 



CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



the equation, then divide both sides by 3. 



3a; + 5 = 14 
3a; + 5-5 = 14 -5 

3a; = 9 
3a: _ 9 
T ~ 3 

x = 3 



Original equation. 

To "undo" adding 5, subtract 5 
from both sides of the equation. 

Simplify both sides. 

To "undo" multiplying by 3, divide 
both sides of the equation by 3. 

Simplify both sides. 



Check: To check the solution, substitute 3 for x in the original equation and 
simplify. 



3a; + 5 = 14 

3(3) + 5= 14 

9 + 5= 14 

14= 14 



Original equation. 
Substitute 3 for x. 
Multiply first: 3(3) = 9. 
Add: 9 + 5 = 14. 



Answer: x = 2 



Because the last line of the check is a true statement, this guarantees that 3 is 
a solution of the original equation. 

□ 



You Try It! 



Solve for x: 

x 3 _ 1 
2 ~ 5 ~ 4 



Let's try an equation with fractions. 



EXAMPLE 2. Solve for 



x 2 _ 1 

5 ~ 3 ~ 2 

Solution: On the left, order of operations demands that we first divide x by 
5, then subtract 2/3. To solve this equation for x, we must "undo" each of 
these operations in inverse order. Thus, we will first add 2/3 to both sides of 
the equation, then multiply both sides of the resulting equation by 5. 



x 2 _ 1 

5 ~ 3 ~ 2 
x 2 2 _ 1 2 



Original equation. 

To "undo" subtracting 2/3, 

add 2/3 to both sides of the equation. 



On the left, we simplify. On the right, we make equivalent fractions with a 
common denominator. 



x 

5 ~ 

x _ 7 

5 ~ 6 



3 4 
6 + 6 



Make equivalent fractions 
3 4 7 



Add: 



6 6 6' 



2.2. SOLVING EQUATIONS: MULTIPLE STEPS 



s<) 



Now we "undo" dividing by five by multiplying both sides of the equation by 
5. 



5 (I) 



35 



Multiply both sides by 5. 



On the left, simplify. On the right, 
multiply: 5 ( -J = —. 



Check: Let's use the TI-84 to check this solution. 

1. Store the value 35/6 in the variable X using the following keystrokes. 



HEBE 



ENTER 



The result is shown in Figure 2.5. 

2. Enter the left-hand side of the original equation: x/5 — 2/3. Use the 
following keystrokes. 



EBEBE 



ENTER 



The result is shown in Figure 2.5). 



5.833333333 
.5 



3. Press the MATH button on your calculator (see Figure 2.6), then select 

l:^Frac, then press the ENTER button. This will convert the decimal Figure 2.5: Checking the so- 
result to a fraction (see Figure 2.6). lution to x/5 - 2/3 = 1/2. 



BjaU HUM CPtt PRE 




35/6-»X 


ija^Frac 




5.S33333333 


Z: ►Dec 




y./5-2/Z 


3:^ 




.5 


4:*J-< 




flns^Frac 


5: *f 




1/2 


6:fMin<: 




I 


7-i-fMaxC 







Figure 2.6: Changing the result to a fraction. 



1 35 x 

Note that the result is — , showing that — is a solution of — 

2 s 6 5 



Answer: x = 17/10 



□ 



Let's try an equation with decimals. 



90 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



You Try It! 



Solve for x: 

3.25- 1.2a; = 0.37 



EXAMPLE 3. Solve for 



5.2a; + 2.3 



-3.94 



Solution: On the left, order of operations demands that we first multiply x 
by 5.2, then add 2.3. To solve this equation for x, we must "undo" each of 
these operations in inverse order. Thus, we will first subtract 2.3 from both 
sides of the equation, then divide both sides by 5.2. 



Original equation. 

2.3 To "undo" adding 2.3, subtract 
2.3 from both sides. 

On the left, simplify. On the right, 
add: -3.94-2.3 = -6.24. 

To "undo" multiplying by 5.2, 
divide both sides by 5.2. 

On the left, simplify. On the right, 
divide: -6.24/5.2 = -1.2. 



5.2a; + 2.3 = 


-3.94 


5.2a; + 2.3 -2.3 = 


-3.94 


5.2a; = 


-6.24 


5.2a; 


-6.24 


5.2 


5.2 


x = 


-1.2 



Check: To check the solution, substitute —1.2 for x in the original equation 
and simplify. 



5.2a; + 2.3 = 


-3.94 


Original equation. 


5.2(-1.2) + 2.3 = 


-3.94 


Substitute —1.2 for x. 


-6.24 + 2.3 = 


-3.94 


Multiply: 5.2(-1.2) = -6.24 


-3.94 = 


-3.94 


Add: -6.24 + 2.3 = -3.94. 



Answer: x = 2.4 



Because the last line of the check is a true statement, this guarantees that —1.2 
is a solution of the original equation. 

□ 



Variables on Both Sides of the Equation 

It is not uncommon that the variable you are solving for appears in terms on 
both sides of the equation. Consider, for example, the equation 2a; + 3 = 5 — 7a;. 
In cases like this, it is helpful to have a general understanding of what it means 
to "solve for a;." 



Solve for x. 


When asked to solve 


an 


equation 


for x, 


the 


goal is 


to 


manipulate 


the equation 


into the final form 

X 


= 


"Stuff", 













2.2. SOLVING EQUATIONS: MULTIPLE STEPS 91 



where "Stuff" is a valid mathematical expression that may contain other vari- 
ables, mathematical symbols, etc., but it must not contain any occurrence of 
the variable x. 



In this section, "Stuff" will always be a single number, but in Section 2.4, 
Formulae, "Stuff" will take on added complexity, including variables other 
than x. 



Strategy for solving for x. When asked to solve an equation for x, a common 
strategy is to isolate all terms containing the variable x on one side of the 
equation and move all terms not containing the variable x to the other side of 
the equation. 



Solution: We need to isolate all terms containing x on one side of the equation. 
We can eliminate 5a; from the right-hand side of 3 — 2x = 5a; + 9 by subtracting 
5a; from both sides of the equation. 

3 — 2x = 5a; + 9 Original equation. 

3 — 2x — 5a; = 5a; + 9 — 5x Subtract 5a; from both sides. 

3 — 7x = 9 Simplify both sides. 

Next, eliminate 3 from the left-hand side of the last equation by subtracting 3 
from both sides of the equation. 

3 — 7a; — 3 = 9 — 3 Subtract 3 from both sides. 

— 7a; = 6 Simplify both sides. 

Note how we have isolated all terms containing x on one side of the equation. 



Divide both sides by —7. 
Simplify both sides. 



-7x 


6 


-7 


^7 


x = 


6 

~7 



You Try It! 



EXAMPLE 4. Solve 3 - 2a; = 5a; + 9 for x. Solve for 



4a; + 7 = 5 - 8a; 



92 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Answer: x = —1/6 



Check: To check the solution, substitute —6/7 for x in the original equation. 

Original equation. 

Substitute —6/7 for x. 

Multiply: -2(-6/7) = 12/7 and 
5(-6/7) = -30/7. 

Make equivalent fractions with 
a common denominator. 



3-2x~- 


= 5a; + 9 


2 B)^ 


-(-?)♦ 


12 

3 + y = 


30 

= -y + 9 


21 12 

y + y = 


30 63 

= -y + y 



Add. 



33 _ 33 

y ~ y 

Because the last line of the check is a true statement, this guarantees that —6/7 
is a solution of the original equation. 

□ 



You Try It! 



Solve for x: 

2x-(x-2) = 2(a; + 7) 



Simplifying Expressions When Solving Equations 

Sometimes we need to simplify expressions before we can isolate terms contain- 



ing x. 



EXAMPLE 5. Solve for x: 2(3x + 1) - 3(4 - 2x) = -34 

Solution: We'll first simplify the expression on the left-hand side of the 
equation using the Rules Guiding Order of Operations. 



2(3a; + l) -3(4 -2x) = -34 
6a; + 2- 12 + 6a; = -34 

12a;- 10 = -34 



Original equation. 

Multiply: 2(3a; + 1) = 6x + 2. 

Multiply: -3(4 - 2a;) = -12 + 6x. 

Add: 6a; + 6a; = \2x. 

Add: 2- 12 = -10. 



To "undo" subtracting 10, we add 10 to both sides of the equation. 

12a; - 10 + 10 = -34 + 10 Add 10 to both sides. 
12a; = — 24 Simplify both sides. 

To "undo" multiplying by 12, we divide both sides by 12. 

-24 



12a; 
x 



12 
-2 



Divide both sides by 12. 
Simplify both sides. 



Check: Let's use the TI-84 to check this solution. 



2.2. SOLVING EQUATIONS: MULTIPLE STEPS 



93 



1. First, store —2 in the variable X using the following keystrokes. 



EDS 



ENTER 



The result is shown in the first image in Figure 2.7. 

2. Enter the left-hand side of original equation: 2(3o; + 1) — 3(4 — 2x). Use 
the following keystrokes. 



2 D > 



□ 







HBH 



ENTER 



The result is shown in the second image in Figure 2.7. 




2*<3*X+l)-3*<4-2 
-34 



Figure 2.7: Checking the solution to 2(3x + 1) - 3(4 - 2x) 



-34. 



Note that when —2 is substituted for x in the left-hand of the equation, the 
result is —34, equalling the right-hand side of the equation. Thus, the solution 
— 2 checks. Answer: x 



12 



□ 



You Try It! 



EXAMPLE 6. Solve for x: 2x - 5(3 - 2x) = A(x - 1) 

Solution: We'll first simplify the expressions on each side of the equation 
using the Rules Guiding Order of Operations. 



Solve for x: 

5(1 -x) = 2(x + 3)- (x- 1) 



2a; -5(3 -2a;) = i(x - 1) 
2x- 15+ 10a; = 4x - 4 

12a;- 15 = 4a; -4 



Original equation. 
On the left, distribute the —5. 
On the right, distribute the 4. 
On the left, add: 2a; + lOx = 12a; 



94 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

Next, we need to isolate terms containing the variable x on one side of the 
equation. To remove the term 4a; from the right-hand side, we subtract Ax 
from both sides of the equation. 

\2x — 15 — 4a; = 4x — 4 — Ax Subtract 4a; from both sides. 

8a; — 15 = —4 Simplify both sides. 

To remove the term —15 from the left-hand side, we add 15 to both sides of 
the equation. 

8a; - 15 + 15 = -4 + 15 Add 15 to both sides. 

8a; = 11 Simplify both sides. 

Finally, to "undo" multiplying by 8, we divide both sides by 8. 

— = — Divide both sides by 8. 

8 8 y 

x = — Simplify both sides. 

8 

Check: Let's use the TI-84 to check this solution. 

1. First, store 11/8 in the variable X using the following keystrokes. 



1 X,T,0,n 1 


I 


ENTER 



□ □HE 



The result is shown in the first image in Figure 2.8. 

2. Enter the left-hand side of the original equation: 2x — 5(3 — 2a;). Use the 
following keystrokes. 



ENTER 



The result is shown in the second image in Figure 2.8. 

3. Enter the right-hand side of the original equation: 4(a; — 1). Use the 
following keystrokes. The result is shown in the third image in Figure 2.8. 



|T| |] Q^ \T\ 



ENTER 



2.2. SOLVING EQUATIONS: MULTIPLE STEPS 



m 



ll/8-»X 



1.375 



2*X-5*<3-2*X) 



1.5 



4*CX-1) 



1.5 



Figure 2.8: Checking the solution to 2x — 5(3 — 2x) = A(x — 1). 



There is no need to use the l:^Frac from the MATH menu this time. The 
fact that both sides of the equation evaluate to an identical 1.5 when x = 11/8 
guarantees that 11/8 is a solution of 2x — 5(3 — 2x) = 4(x — 1). 



Answer: 



■1/3 



□ 



96 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



t* ■ ;-*■ ;*■ 



Exercises 



■*:>•*'">■*; 



In Exercises 1-16, solve the given equation for x. 



1. 2a;-20 = -12 

2. -Ax- 1 =3 

3. -ll + 3a; = -44 

4. -8+Ux = -22 

5. -5a; + 17= 112 

6. 3a; + 12 = 51 

7. -16a;- 14 = 2 

8. 4a; - 4 = 64 



9.5- 13a; = 70 

10. 11 + 10a; = 81 

11. -2-9a; = -74 

12. -4- 16a; = -100 

13. 7- x = -7 

14. 20 - 3a; = 35 

15. -Ax + 14 = 74 

16. -4a; +15 = 27 



In Exercises 17-24, solve the equation and simplify your answer. 



17. 



18. 



19. 



x 



1 _ 9 

7 ~ 3 ~ ~8 
x 8 _ 4 

8 ~ 3 ~ ~7 
a; 4 _ 3 

7 + 9 ~ 2 



x 1 8 
20. - + - = - 

5 6 7 



21. 


a; 2 4 
2 + 3 ~ 7 


22. 


a; 4 3 

7 + 5 ~ 4 


23. 


a; 9 5 
5 ~ 2 ~ ~3 


24. 


a; 8 3 
5 ~ 9 ~ ~2 



In Exercises 26-32, solve each equation. 



25. 0.3a; + 1.7 = 3.05 

26. -7.2a; + 2.9 = 64.10 

27. 1.2a; + 5.2 = 14.92 

28. -7.3a; +1.8 = -45.65 



29. 3.5a; -3.7= -26.10 

30. -1.4a; -4.7 = 5.80 

31. -4.7a;- 7.4 = -48.29 

32. -5.2a;- 7.2 = 38.04 



In Exercises 33-44, solve each equation. 



33. 13-9a; = 11 - 5a; 



34. 11- 10a; = 13 - 4a; 



2.2. SOLVING EQUATIONS: MULTIPLE STEPS 



97 



35. 11a; + 10= 19a; + 20 

36. 20a; + 19 = 10a; + 13 

37. 11- 15a; = 13- 19a; 

38. 13- 11a; = 17 -5a; 

39. 9a; + 8 = 4- 19a; 



40. 10a; + 8 = 6 - 2a; 

41. 7a; + 11 = 16- 18a; 

42. llx + 8 = 2- 17a; 

43. 12a; + 9 = 4a; + 7 

44. 6a; + 3 = 16a; + 11 



In Exercises 45-56, solve each equation. 

45. 8(5x-3) -3(4a; + 6) =4 

46. 6(3a;-8) - 6(4ai + 6) = 3 

47. 2a; - 4(4 - 9a;) = 4(7a; + 8) 

48. 4a; - 9(6 - 2x) = 2(5x + 7) 

49. 2(6 - 2x) - (4a; - 9) = 9 

50. 2(8 -5a;)- (2a; - 6) = 4 



51. 3(5x-6)- 7(7a; + 9) = 3 

52. 9(3a; - 7) - 9(2a; + 9) = 6 

53. 2a; - 2(4 - 9a;) = 8(6a; + 2) 

54. 3a; - 3(5 - 9a;) = 6(8a; + 2) 

55. 2(7 -9a;)- (2a; - 8) = 7 

56. 8(5 -2a;)- (8a; - 9) = 4 



;*•;-*■ i± 



Answers 



•*;■*■; ■*'> 



1. 


4 


3. 


-11 


5. 


-19 


7. 


-1 


9. 


-5 


11. 


8 


13. 


14 


15. 


-15 


17. 


13S 

~~24 




133 



21. 



21 



23. 


85 


25. 


4.5 


27. 


8.1 


29. 


-6.4 


31. 


8.7 




1 



33. 



35. 



37. 



19. 



18 



98 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



39. 


1 

~7 


41. 


1 
5 


43. 


1 

~4 


45. 


23 
14 


47. 


24 

y 



49. 


3 

2 


51. 


42 
~17 


53. 


6 

~7 


55. 


3 

4 



2.3. CLEARING FRACTIONS AND DECIMALS 99 



2.3 Clearing Fractions and Decimals 

In this section we introduce techniques that clear fractions and decimals from 
equations, making the resulting equation a lot easier to solve. 

When clearing fractions from an equation, you will need to simplify products 
like the ones posed in the following examples. 



You Try It! 



EXAMPLE 1. Simplify: 12 (1 X) . ^^ 

Solution: When we multiply three numbers, such as 12, 2/3, and x, the /$ 

associative property of multiplication tells us that it does not matter which \ T x 

two numbers we multiply first. We use the associative property to regroup, 
then multiply numerators and denominators and simplify the result. 

(2 \ ( 2\ 
12 —x 1 = I 12 • — 1 x Associative property of multiplication. 

24 



-x Multiply: 12 ■ 2 = 24. 

o 

8x Divide: 24/3 = 8. 



Answer: 9x 

□ 



Example 1 shows all of the steps involved in arriving at the answer. How- 
ever, the goal in this section is to perform this calculation mentally. So we just 
"Multiply 12 and 2 to get 24, then divide 24 by 3 to get 8." This approach 
allows us to write down the answer without doing any work. 

12 (-x) =8x 

You should practice this mental calculation until you can write down the answer 
without writing down any steps. 



You Try It! 



EXAMPLE 2. Simplify: 18 ( -x , 

Solution: This time we perform the calculations mentally. Multiply 18 and 2 /g 

to get 36, then divide 36 by 9 to get 4. 14 I —x 

18 (§*) = Ax 

Answer: 6x 



□ 



100 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

When the numbers get larger, the mental calculations become harder. For 
example, consider 



72 , x 



In this case, the work "multiply 72 and 8 to get 576, then divide 576 by 9 to get 
64" is a bit difficult to carry in your head. However, this is when the calculator 
comes to the rescue. 



You Try It! 



Use your calculator to 
simplify: 



l r 



EXAMPLE 3. Use your calculator to help simplify 72 ( -x 



Solution: Use your calculator to multiply 72 and 8, then divide by 9. Enter 
72*8/9 and press the ENTER key. 



72*3^9 



64 



Answer: 45a; 



Thus, 72 [ -x 1 = 64cc. 



□ 



Canceling is More Efficient 

In Examples 1, 2, and 3, we multiplied numerators, then divided by the sole 
denominator. We also saw that it is a bit difficult to carry the work in our head 
as the numbers grow larger. In Chapter 1, Section 3, we saw that canceling 
reduces the size of the numbers and simplifies the work. 



You Try It! 



Simplify: 



EXAMPLE 4. Simplify: 72 ( -x 



64 1 -x 



Solution: In Example 3, we used our calculator to multiply 72 and 8 to get 
576, then divided 576 by 9 to get 64. In this solution, we divide 9 into 72 to 
get 8, then multiply 8 by 8 to get 64. We get the same answer, but because 
the intermediate numbers are much smaller, the calculations are much easier 
to do mentally. 



2.3. CLEARING FRACTIONS AND DECIMALS 



101 



72I-* 



72 • - )x 
9, 



(8-8)x 
64cc 



Associative property of multiplication. 



Divide: 72/9 
Multiply: 8 ■ I 



64. 



Answer: 4Cte 



Example 4 shows all of the steps involved in arriving at the answer. Again, 
the goal in this section is to perform this calculation mentally, so we just 
"Divide 9 into 72 to get 8, then multiply 8 by 8 to get 644." 



□ 



72,-* 



64x 



Not only does this approach allow us to write down the answer without doing 
any work, the numerical calculations involve smaller numbers. You should 
practice this mental calculation until you can write down the answer without 
writing down any steps. 



You Try It! 



EXAMPLE 5. Simplify: 27 I -x 

Solution: Divide 9 into 27 to get 3, then multiply 3 by 5 to get 15. 

27 (-a; | = I'm 



Simplify: 



l l X 



Answer: 27a; 



Note: The technique shown in Examples 4 o-nd 5 is the technique we '11 use in 
the remainder of this section. Dividing (canceling) first is far more efficient, 
the smaller numbers allowing us to perform the calculation mentally. 

Clearing Fractions from an Equation 

Now that we've done the required fraction work, we can now concentrate on 
clearing fractions from an equation. Once the fractions are removed from the 
equation, the resulting equivalent equation is far easier to solve than the orig- 
inal. 



□ 



Clearing fractions from an equation. To clear fractions from an equation, 
multiply both sides of the equation by the least common denominator. 



102 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



You Try It! 



Solve for x: 



3 _ 1 

4 ~~ 2 



EXAMPLE 6. Solve for 



2 _ 1 

3 ~ 2' 



Solution: The common denominator for 2/3 and 1/2 is 6. We begin by 
multiplying both sides of the equation by 6. 



x 



2 _ 1 

3 ~ 2 



X+ l)= 6 {2 



°' + ' 5 «i)= 6 U 



Original equation. 
Multiply both sides by 6. 

On the left, distribute the 6. 



To simplify 6(2/3), divide 6 by 3 to get 2, then multiply 2 by 2 to get 4. Thus, 
6(2/3) = 4. Similarly, 6(1/2) = 3. 



Qx + 4 = 3 



1 1) =4 > 6 u) 



Note that the fractions are now cleared from the equation. To isolate terms 
containing x on one side of the equation, subtract 4 from both sides of the 
equation. 



6x + 4-4 = 3-4 
6a; = -1 



Subtract 4 from both sides. 
Simplify both sides. 



To "undo" multiplying by 6, divide both sides by 6. 



6x _ -1 

~6~ ~ ir 

1 

X = ~6 



Divide both sides by 6. 
Simplify both sides. 



Check: Let's use the TI-84 to check the solution. 

1. Store —1/6 in the variable X using the following keystrokes. 



EDQHH 



ENTER 



The result is shown in the first line in Figure 2.9. 

2. Enter the left-hand side of the original equation: x + 2/3. Use the fol- 
lowing keystrokes. 



2.3. CLEARING FRACTIONS AND DECIMALS 



103 



X,T,0,n 



0B0 



ENTER 



The result is shown in the second line in Figure 2.9. 

3. Press the MATH key, then select l:^Frac and press the ENTER key. The 
result is shown in the third line in Figure 2.9. 



-l/6-»X 
flns^Frac 



.5 
1/2 



Figure 2.9: Checking that —1/6 is a solution of x + 2/3 = 1/2. 



The result is identical to the right-hand side of the equation x + 2/3 = 1/2. 

Thus, the solution checks. Answer: x = 5/4 



□ 



EXAMPLE 7. Solve for x: 



4 

— x 
5 



You Try It! 



Solve for 



Solution: The common denominator for 4/5 and —4/3 is 15. We begin by 
multiplying both sides of the equation by 15. 



l.-,||, 



15 



Original equation. 
Multiply both sides by 15. 



To simplify 15(4/5), divide 5 into 15 to get 3, then multiply 3 by 4 to get 12. 
Thus, 15(4/5) = 12. Similarly, 15(-4/3) = -20 

12a; = -20 Multiply. 

To "undo" multiplying by 12, we divide both sides by 12. 

Divide both sides by 12. 
Reduce to lowest terms. 



12a; 


-20 


12 


12 




5 


x = 


~3 



3 3 

"7 X= 2 



104 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Check: To check, substitute —5/3 for x in the original equation. 



5 
5 
3 

20 

~15 

4 



Original equation. 
Substitute —5/3 for x. 

Multiply numerators and denominators. 
Reduce. 



Answer: x 



-7/2 



The fact that the last line is a true statement guarantees the —5/3 is a solution 

4 4 

of the equation —x = . 

4 5 3 

□ 



You Try It! 



Solve for 



5x 2 _ 5 3x 
~9~ ~ 3 ~ 9 ~ T 



EXAMPLE 8. Solve for x: —-- = --—. 

3 4 2 4 

Solution: The common denominator for 2a;/3, —3/4, 1/2, and — 3x/4 is 12. 

We begin by multiplying both sides of the equation by 12. 

2x 3 1 3a; 



12 



12 



3 4 
2x 3 
Y~ 4 

12 ■ 3 



2 
12 

12 



4 

1 3a; 

2 ~ T 
1 



12 



Original equation. 
Multiply both sides by 12. 
Distribute the 12 on each side. 



To simplify 12(2a;/3), divide 3 into 12 to get 4, then multiply 4 by 2a; to get 8a;. 
Thus, 12(2x/3) = 8x. Similarly, 12(3/4) = 9, 12(1/2) = 6, and 12(3x/4) = 9a;. 



8x - 9 = 6 - 9x 



Multiply. 



Note that the fractions are now cleared from the equation. We now need to 
isolate terms containing x on one side of the equation. To remove the term 
—9a; from the right-hand side, add 9a; to both sides of the equation. 



8a; - 9 + 9a; = 6 - 9a; + 9a; 
17a; - 9 = 6 



Add 9a; to both sides. 
Simplify both sides. 



To remove the term —9 from the left-hand side, add 9 to both sides of the 
equation. 



17a;- 9 + 9 = 6 + 9 
17a; = 15 



Add 9 to both sides. 
Simplify both sides. 



2.3. CLEARING FRACTIONS AND DECIMALS 



105 



Finally, to "undo" multiplying by 17, divide both sides of the equation by 17. 



17a: _ 15 

TT ~ 17 
15 

17 



X 



Divide both sides by 17. 
Simplify both sides. 



Check: Let's use the TI-84 to check the solution. 

1. Store the number 15/17 in the variable X using the following keystrokes. 



Q0BQQ 



X,T,0,n I ENTER 



The result is shown in Figure 2.10. 



.8823529412 



Figure 2.10: Storing 15/17 in X. 



2x 3 

2. Enter the left-hand side of the original equation: . Use the fol- 

S H 3 4 

lowing keystrokes. 







QBQBQ 



ENTER 



The result is shown in the first image in Figure 2.11). 

1 3a; 
3. Enter the right-hand side of the original equation: — . Use the 

following keystrokes. 



□ BEBH 



□ 



ENTER 



The result is shown in the second image in Figure 2.11. 

Because both sides simplify to —.1617647059 when 15/17 is substituted for x, 

this guarantees that 15/17 is a solution of the equation 2x/3 — 3/4 = 1/2 — 3a;/4. Answer: x = 22/37 



□ 



106 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



2*X/3-3/4 

-.1617647059 



l/2-3*X/4 

-.1617647059 



Figure 2.11: Checking 15/17 in 2a;/3 - 3/4 = 1/2 - 3x/4. 



Clearing Decimals from an Equation 

Multiplying by the appropriate power of ten will clear the decimals from an 
equation, making the resulting equivalent equation far easier to solve. 

Before we begin, first recall the following facts about multiplying by powers 
of ten. 



You Try It! 



Solve for x: 

1.34- 4.5a; = 2.2 



• 10(1.2345) = 12.345. Multiplying by 10 moves the decimal point one 
place to the right. 

• 100(1.2345) = 123.45. Multiplying by 100 moves the decimal point two 
places to the right. 

• 1000(1.2345) = 1234.5. Multiplying by 1000 moves the decimal point 
three places to the right. 

Note the pattern: The number of zeros in the power of ten determines the 
number of places to move the decimal point. So, for example, if we multiply 
by 1,000,000, which has six zeros, this will move the decimal point six places 
to the right. 



EXAMPLE 9. Solve for 



2.3a; - 1.25 = 0.04a;. 



Solution: The first term of 2.3a; — 1.25 = 0.04a; has one decimal place, the 
second term has two decimal places, and the third and final term has two 
decimal places. At a minimum, we need to move each decimal point two places 
to the right in order to clear the decimals from the equation. Consequently, we 
multiply both sides of the equation by 100. 



2.3a;- 1.25 

100(2. 3a;- 1.25) 

100(2. 3a;) - 100(1.25) 

230a; - 125 



0.04a; Original equation. 

100(0.04x) Multiply both sides by 100. 
100(0.04a;) Distribute the 100. 
4a; Multiplying by 100 moves all decimal 

points two places to the right. 



2.3. CLEARING FRACTIONS AND DECIMALS 107 

Note that the decimals are now cleared from the equation. We must now isolate 
all terms containing x on one side of the equation. To remove the term Ax from 
the right-hand side, subtract Ax from both sides of the equation. 

230a; — 125 — Ax = Ax — Ax Subtract Ax from both sides. 

226a; - 125 = Simplify. 

To remove —125 from the left-hand side, add 125 to both sides of the equation. 

226a; - 125 + 125 = + 125 Add 125 to both sides. 

226a; = 125 Simplify both sides. 

Finally, to "undo" multiplying by 226, divide both sides by 226. 

226a: 125 

Divide both sides by 226. 

Simplify. 

Check: Let's check the answer with the TI-84. 

1. Store 125/226 in the variable X using the following keystrokes. 



226 


226 


x = 


125 

226 



Q00H000 



ENTER 



The result is shown in the first image in Figure 2.12. 

2. Enter the left-hand side of the original equation: 2.3a: — 1.25. Use the 
following keystrokes. 



0D0I BBB0000 



ENTER 



The result is shown in the second image in Figure 2.12. 

3. Enter the right-hand side of the original equation: 0.04a;. Use the follow- 
ing keystrokes. 

mnmEDBB 

The result is shown in the third image in Figure 2.12. 



108 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



125/226-*X 

.5530973451 



2.3*X-1.25 

.0221238938 



0.04*X 



,0221238938 



Figure 2.12: Checking 125/226 in 2.3a; - 1.25 = 0.04s. 



Answer: x = -43/225 



Note that both sides yield the same decimal approximation 0.0221238938 when 
125/226 is substituted for x. This guarantees that 125/226 is a solution of 
2.3a;- 1.25 = 0.04a;. 



□ 



2.3. CLEARING FRACTIONS AND DECIMALS 



109 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



In Exercises 1-6, simplify the expression. 



9 



1. !.(.( -.t 



2. 27 ( -a; 



3. 14 | -x 



4. -12 ( -a; 



5. 70 ( -x 



6. -27 [ -x 



In Exercises 7-18, for each of the following equations, clear fractions by multiplying both sides by the 
least common denominator. Solve the resulting equation for x and reduce your answer to lowest terms. 



7. 


9 1 

~7 X ~ 3 


5 
~ 3 


8. 


1 3 

~2 X ~ 4 


5 
~ ~9 


9. 


7 5 
-x+ - = 
3 9 


2 4 
— a; 

3 3 


10. 


2 9 

3^" 4 ~ 


5 4 
"S^" 3 


11. 


9 8 
4 X "7 = 


3 

2 


12. 


7 2 
~3 X ~ 9 


4 
~3 



13. 





4 3 






14. 


2 5 
"j^" 7 






15. 


3 6 

4 5 






16. 


2 1 

X ~ 9 ~~ 4 






17. 


1 4 
"j}^" 3 " 


3 


8 
" 5 



, o 6 3 9 1 

18. — a; = — x 

7 5 7 2 



In Exercises 19-32, clear decimals from the given equation by multiplying by the appropriate power of 
ten, then solve the resulting equation for x. Your final answer should be a fraction reduced to lowest 
terms. 



19. 2.39a; + 0.71 = -1.98a; + 2.29 

20. 0.12a; + 0.52 = -1.47a; - 2.12 

21. 0.4a;- 1.55 = 2.14 

22. 0.8a; -2.18 = 1.49 

23. 2.6a; -2.54 = -2.14a; 

24. -1.4a; -2.98 = 0.55a; 

25. 0.7a; = -2.3a; - 2.8 



26. 3.4a; = 1.8s + 2.5 

27. -4.8a; -2.7= -1.9 

28. -2.4a; + 2.5 = 2.3 

29. 1.7a; + 2.1 = -1.6a; + 2.5 

30. -1.2a; + 0.4 = -2.7a; - 1.9 

31. 2.5a; + 1.9 = 0.9a; 

32. 4.4a; + 0.8 = 2.8x 



1. 


72a; 


3. 


21a; 


5. 


90a: 


7 


14 




9 


q 


17 




15 


11. 


74 
63 


13. 


32 


15. 


9 
20 


17. 


16 
~25 



110 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



m- j*- j*- Answers •*$ •** •** 



19. 


158 
437 


21. 


369 


23. 


127 
237 


25. 


14 
~15 


27. 


1 

~6 


29. 


4 
3.3 


31. 


19 
~16 



2.4. FORMULAE 



111 



2.4 Formulae 

The formulae of science usually contain variable letters other than the variable 
x. Indeed, formulae in science typically use several letters. For example, Isaac 
Newton's Universal Law of Gravitation says that the magnitude of the force of 
attraction between two celestial bodies is given by the formula 

GMm 



where inn usually denotes the mass of the smaller body, M the mass of the larger 
body, and r is the distance between the two bodies. The letter G represents 
the universal gravitational constant, having value 6.67428 x 10~ n N(m/kg) . 

Variable case. Note the use of upper and lower case letter M's in Newton's 
Law of Gravitation. When working with scientific formulae, you must maintain 
the case of the given letters. You are not allowed to substitute lower for upper 
case, or upper for lower case in your work. 

In Section 2.2, we described the goal that must be met when we are asked 
to "solve an equation for ic." 

Solve for x. When asked to solve an equation for x, the goal is to manipulate 
the equation into the final form 

x= "Stuff", 

where "Stuff" is a valid mathematical expression that may contain other vari- 
ables, mathematical symbols, etc., but it must not contain any occurrence of 
the variable x. 



'Formulae" is the plural for 
'formula." 



Thus, to solve an equation for x, we need to isolate the terms containing x 
on one side of the equation, and all remaining terms on the other side of the 
equation. 



EXAMPLE 1. Solve for a;: x + a = b. 

Solution: To undo the effects of adding a, subtract a from both sides of the 
equation. 



x + a = b 
x + a — a = b — a 
x = b — a 



Original equation. 
Subtract a from both sides. 
Simplify. 



You Try It! 



Solve for x: 

x — c = d 

Answer: x = c + d. 



□ 



You Try It! 



Solve for x: 

x — c = d 



Answer: c = x — d. 



112 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

In Example 1, note that the answer x = b — a has the required form, x = 
"Stuff" , where "Stuff" is a valid mathematical expression that contains other 
variables, mathematical symbols, etc., but it does not contain any occurrence 
of the variable x. Now, what if we were asked to solve the same equation for 
a, instead of xl 



EXAMPLE 2. Solve for a: x + a = b. 

Solution: We are instructed to solve the equation x + a = b for a. This means 
that our final answer must have the form a = "Stuff" , where "Stuff" is a valid 
mathematical expression that contains other variables, mathematical symbols, 
etc., but it does not contain any occurrence of the variable a. This means that 
we must isolate all terms containing the variable a on one side of the equation, 
and all remaining terms on the other side of the equation. Now, to undo the 
effect of adding x, subtract x from both sides of the equation. 



x + a = b 
x + a — x = b — x 
a = b — x 

Note that we have a = "Stuff" 
variable we are solving for. 



Original equation. 
Subtract x from both sides. 
Simplify. 

where "Stuff" contains no occurrence of a, the 



a 



You Try It! 



Solve for m: 



E = mc 



EXAMPLE 3. The formula F = kx, known as "Hooke's Law" , predicts the 
force F required to stretch a spring x units. Solve the equation for k. 

Solution: We are instructed to solve the equation F = kx for k. This means 
that our final answer must have the form k = "Stuff" , where "Stuff" is a 
valid mathematical expression that may contain other variables, mathematical 
symbols, etc., but it may not contain any occurrence of the variable k. This 
means that we must isolate all terms containing the variable k on one side 
of the equation, and all remaining terms on the other side of the equation. 
However, note that all terms containing the variable k are already isolated on 
one side of the equation. Terms not containing the variable k are isolated on 
the other side of the equation. Now, to "undo" the effect of multiplying by x, 
divide both sides of the equation by x. 

Original equation. 

Divide both sides by x. 

Simplify. 



F = 


kx 


F 


kx 


X 


X 


F 





2.4. FORMULAE 



113 



Saying that F/x = k is equivalent to saying that k = F/x. We can leave our 
answer in the form shown in the last step, but some instructors insist that we 
write the answer as follows: 



k 



F 
x 



F/x = k is equivalent to k = F/x. 



Note that we have k = "Stuff" , where "Stuff" contains no occurrence of k, the 

variable we are solving for. Answer: m 



E 



□ 



EXAMPLE 4. The formula V = RI is called "Ohm's Law." It helps calculate 
the voltage drop V across a resistor R in an electric circuit with current /. Solve 
the equation for R. 

Solution: We are instructed to solve the equation V = RI for R. This means 
that our final answer must have the form R = "Stuff" , where "Stuff" is a 
valid mathematical expression that may contain other variables, mathematical 
symbols, etc., but it may not contain any occurrence of the variable R. This 
means that we must isolate all terms containing the variable R on one side 
of the equation, and all remaining terms on the other side of the equation. 
However, note that all terms containing the variable R are already isolated on 
one side of the equation. Terms not containing the variable R are isolated on 
the other side of the equation. Now, to "undo" the effect of multiplying by I, 
divide both sides of the equation by /. 



V = RI 

V _RI 

7 ~ T 

v 

7 



R 



Original equation. 
Divide both sides by I. 

Simplify. 



This can also be written in the following form: 



R 



V 

7 



v/i. 



Note that we have R = "Stuff 
the variable we are solving for. 



V/I = R is equivalent to R 

' , where "Stuff" contains no occurrence of 7?, 



You Try It! 



Solve for t: 



si 



Answer: t 



d 

s 



□ 



Clearing Fractions 

If fractions occur in a formula, clear the fractions from the formula by multi- 
plying both sides of the formula by the common denominator. 



114 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



You Try It! 



Solve for g: 



s = -gV 



EXAMPLE 5. The formula K = —mv 2 is used to calculate the kinetic energy 
K of a particle of mass m moving with velocity v. Solve the equation for m. 

Solution: We're asked to solve the equation K = (l/2)mv 2 for m. First, clear 
the fractions by multiplying both sides by the common denominator. 



K = —mv 



2{K) = 2 ( -mv 2 
2K = mv 2 



Original equation. 

Multiply both sides by 2. 
Simplify. Cancel 2's. 



Note that all terms containing m, the variable we are solving for, are already 
isolated on one side of the equation. We need only divide both sides by v 2 to 
complete the solution. 



2s 
Answer: q = -r- 

t 2 



2K mv 2 

v 2 

2K 

?.' 2 



v 
m 



Divide both sides by v . 
Simplify. Cancel v for v . 



Note that the final answer has the form m = "Stuff" , where "Stuff" contains 
no occurrence of the variable m. 

□ 



You Try It! 



Solve for q^. 



F 



kqiq 2 



EXAMPLE 6. As mentioned earlier, Newton's Universal Law of Gravitation 
is described by the formula 

„ GMm 



Solve this equation for m. 

Solution: We're asked to solve the equation F = GMm/r 2 for m. First, clear 
the fractions by multiplying both sides by the common denominator. 



GMr 



w-*(2p) 



r 2 F = GMm 



Original equation. 

Multiply both sides by r . 
Simplify. Cancel r for r . 



2.4. FORMULAE 



115 



Note that all terms containing m, the variable we are solving for, are already 
isolated on one side of the equation. We need only divide both sides by GM 
to complete the solution. 



r 2 F GMr 



GM 
r 2 F 
GM 



GM 



m 



Divide both sides by GM. 
Simplify. Cancel GM for GM. 



Note that the final answer has the form m 
no occurrence of the variable m. 



"Stuff" , where "Stuff" contains 



Answer: </2 = ~, 



Fr 2 
kqi 



□ 



Geometric Formulae 

Let's look at a few commonly used formulae from geometry. 



EXAMPLE 7. Let W and L represent the width and length of a rectangle, 
respectively, and let P represent its perimeter. 



You Try It! 



The perimeter of a rectangle 
is 160 meters and its width is 
30 meters. Finds its length. 



W 



IT" 



The perimeter (distance around) of the rectangle is found by summing its four 
sides, then combining like terms. 



p=L+W+L+W 
P = 2W + 2L 



Summing the four sides. 
Combine like terms. 



Problem: Solve P = 2W + 2L for L. Then, given that the perimeter is 300 
feet and the width is 50 feet, use your result to calculate the length. 



Solution: We're first asked to solve P = 2W + 2L for L. First, isolate all 



116 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



terms containing the variable L on one side of the equation. 

Original equation. 

Subtract 2W from both sides. 

Simplify. 

Divide both sides by 2. 

Simplify. 

z 

Note that the final result has L = "Stuff', where "Stuff" contains no occurrence 
of the variable L. 

The second part of this example requests that we find the length of the 
rectangle, given that the perimeter is P = 300 feet and the width is W = 50 
feet. To calculate the length, substitute P = 300 and W = 50 in L = [P — 
2W)/2. 





p 


= 2W + 2L 




p 


-2W 


= 2W + 2L- 


-2W 


p 


-2W 


= 2L 




p - 


-2W 


2L 




p- 


2 
-2W 


2 
- L 





P-2W 



L = 


2 




L = 


300- 


2(50) 




2 


L = 


300- 


100 


2 




L = 


200 
~2~ 




L = 


100 





Perimeter formula solved for L. 

Substitute 300 for P, 50 for W. 

Multiply: 2(50) = 100. 

Subtract: 300 - 100 = 200. 
Divide: 200/2 = 100. 



Answer: L = 50 meters 



Hence, the length of the rectangle is 100 feet. 



□ 



You Try It! 



The area of a triangle is EXAMPLE 8. Let 6 and h represent the length of the base and the height 

140 cm 2 and the length of its of a triangle, respectively, and let A represent the area of the triangle, 
base is 70 cm. Find the 
height of the triangle. 

h 




The area of the triangle is computed using the formula: 

A = -bh 

2 



2.4. FORMULAE 117 

That is, the area of a triangle is "one-half the base times the height." 

Problem: Solve the formula A = ^bh for h. Secondly, given that the area is 
A = 90 in" (90 square inches) and the length of the base is 15 in (15 inches), 
find the height of the triangle. 

Solution: We're first asked to solve A = (l/2)bh for h. Because the equation 
has fractions, the first step is to clear the fractions by multiplying both sides 
by the least common denominator. 

A = — bh Area of a triangle formula. 

2 

2(A) = 2 ( - bh ) Multiply both sides by 2. 

2A = bh Simplify. 

Now, we already have all terms containing the variable h on one side of the 
equation, so we can solve for h by dividing both sides of the equation by b. 

2A bh 

— = — Divide both sides by o. 

b b 

—— = h Simplify. 

Note that the final result has h = "Stuff', where "Stuff" contains no occurrence 
of the variable h. 

The second part of this example requests that we find the height of the 
triangle, given that the area is A = 90 in and the length of the base is b = 15 in. 
To calculate the height of the triangle, substitute A = 90 and b = 15 in 
h = 2A/b. 

94 

Area formula solved for h. 

Substitute 90 for A, 15 for b. 

Multiply: 2(90) = 180. 
Divide: 180/15 = 12. 

Hence, the height of the triangle is 12 inches. Answer: 4 cm 



h = 


2A 
~b 


h = 


2(90) 


15 


h = 


180 


h = 


12 



□ 



118 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



;».;-». t%.. 



Exercises 



•*;■*; -*i 



In Exercises 1-30, solve the given formulas for the indicated variable. 



1. F = kx for x 

2. A = irr 2 for ir 
mc 2 for ?7, 

4. v = Vq + at for vq 

5. A = -nr\ri for r 2 

6. y = mx + b for b 

7. F = ma for a 

8. V = Iwh for I 

9. C = 2irr for r 

10. F = kx for x 

11. y = mx + b for x 

V 

12. I = — for i? 

ii 

13. F = gwB for i> 

14. x + <Z = e for a; 

1 



15. V 



'h for /i 



16. P = IRT for I 

V 

17. I = - for 7? 



18. F = -C + 32 for C 

5 



19. F 



— =- for g 



20. Ax + By = C for y 

21. P = 2W + 2L for VF 

22. A = -h(bt + b 2 ) for h 



\ h ^ 



a + b + c 



23. A 

24. A: 

25. y- 

26. A: 

27. F 

28. A: 

3 

29. d = uf for u 

30. x + d = e for d 



for ft, 



3 








y = m(x - 


xo) 


for 


m 


—bh for b 
2 








GA f for M 






a + b + c 


or h 







31. Let W and L represent the width and 
length of a rectangle, respectively, and let 
A represent its area. 



W 



L 



W 



The area of the rectangle is given by the 
formula 

A = LW. 

Solve this formula for L. Then, given that 
the area of the rectangle is A = 1073 
square meters and its width is W = 29 
meters, determine its length. 

32. Let b\ and b 2 represent the parallel bases 
of a trapezoid and let h represent its 
height. 



2.4. FORMULAE 



119 



The area of the trapezoid is given by the 
formula 



i(* 



b 2 )h. 



Solve this formula for b\. Then, given that 
the area is A = 2457 square centimeters, 
the second base is b 2 = 68 centimeters, 
and the height is h = 54 centimeters, find 
the length b\ of the first base. 

33. A parallelogram is a quadrilateral (four- 
sided figure) whose opposite sides are par- 
allel. 



The area of the parallelogram is computed 
using the formula: 



bh 



Solve this formula for b. Next, given that 
the area is A = 2418 square feet and the 
height is h = 31 feet, find the length of 
the base of the parallelogram. 

34. Let b\ and b 2 represent the parallel bases 
of a trapezoid and let h represent its 
height. 



35 



The area of the trapezoid is given by the 
formula 

A=^(b 1 +b 2 )h. 

Solve this formula for h. Then, given that 
the area is A = 3164 square yards, the 
bases are b\ = 38 yards and 62 = 75 yards, 
find the height h of the trapezoid. 

Let b and h represent the length of the 
base and the height of a triangle, respec- 
tively, and let A represent the area of the 
triangle. 




The area of the triangle is computed using 
the formula: 

A=-bh 
2 

Solve this formula for b. Next, given that 
the area is A = 1332 square inches and the 
height is h = 36 inches, find the length of 
the base of the triangle. 

36. The circumference of a circle, somewhat 
like the term perimeter, is the distance 
around the circle. The diameter of a cir- 
cle is a line segment drawn through the 
center of the circle. 




Since the time of the ancient Greeks, it 
has been known that the ratio of the cir- 
cumference to the diameter is a constant, 
denoted by the symbol w. 

C 

7 = 7r 



120 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Solve the formula for d. Then, given 
C = 188.4 yards and tt = 3.14, find the 
length of the diameter d. 

37. Let W and L represent the width and 
length of a rectangle, respectively, and let 
P represent its perimeter. 
L 



W 



W 



L 



The perimeter (distance around) of the 
rectangle is found by summing its four 
sides, then combining like terms. 

P = 2W + 2L 

Solve P = 2W + 2L for W. Then, given 
that the perimeter is P = 256 meters and 
the length is L = 73 meters, use your re- 
sult to calculate the width. 



j*- j*- j*- Answers •*$-*$■*$ 



F 



E 
3. m = — ; 
cr 



19. q 



21. W 



r 2 F 
~kQ 

P-2L 



5. r 2 



7. a 



F 
m 



9. r = — 

2tt 



11. x 



13. v 



15. h 



17. R 



y-b 
m 

F 
qB 

W 

■KT 2 
V 



23. h 



25. m 



27. M : 



2A 



6i + 6 2 

_ y-yo 

X — Xo 

r 2 F 



G 



111 



d 

29. v = - 

t 



31. 37 square meters 
33. 78 feet 
35. 74 inches 
37. 55 meters 



2.5. APPLICATIONS 



121 



2.5 Applications 

The solution of a word problem must incorporate each of the following steps. 

Requirements for Word Problem Solutions. 

1. Set up a Variable Dictionary. You must let your readers know what 

each variable in your problem represents. This can be accomplished in a 
number of ways: 

• Statements such as "Let P represent the perimeter of the rectangle." 

• Labeling unknown values with variables in a table. 

• Labeling unknown quantities in a sketch or diagram. 

2. Set up an Equation. Every solution to a word problem must include a 

carefully crafted equation that accurately describes the constraints in the 
problem statement. 

3. Solve the Equation. You must always solve the equation set up in the 

previous step. 

4. Answer the Question. This step is easily overlooked. For example, the 

problem might ask for Jane's age, but your equation's solution gives the 
age of Jane's sister Liz. Make sure you answer the original question 
asked in the problem. Your solution should be written in a sentence with 
appropriate units. 

5. Look Back. It is important to note that this step does not imply that 

you should simply check your solution in your equation. After all, it's 
possible that your equation incorrectly models the problem's situation, so 
you could have a valid solution to an incorrect equation. The important 
question is: "Does your answer make sense based on the words in the 
original problem statement." 



Let's give these requirements a test drive. 



You Try It! 



EXAMPLE 1. Three more than five times a certain number is —62. Find 27 more than 5 times a 
the number. certain number is —148. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 



1. Set up a Variable Dictionary. Let x represent the unknown number. 

2. Set up an Equation. "Three more than five times a certain number is —62" 
becomes: 



122 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Three 



more than 



five times 
a certain 
number 

5.T 



-62 



-62 



3 + 5x = 


-62 




+ 5x - 3 = 


-62- 


-3 


5x = 


-65 




5x 


-65 




~~5 ~ 


5 




x = 


-13 





Answer: 



-35 



3. Solve the Equation. To solve for x, first subtract 3 from both sides of the 
equation. 

Original equation. 
Subtract 3 from both sides. 
Simplify. 

Divide both sides by 5. 
Simplify. 

4. Answer the Question. The unknown number is —13. 

5. Look Back. Compute "three more than five times —13." 

3 + 5(-13) = 3 + (-65) 
= -62 

Hence, three more than five times —13 is —62, as required. Our solution 
is correct. 



□ 



You Try It! 



The sum of three consecutive 
odd integers is —225. What 
are the integers? 



EXAMPLE 2. The sum of three consecutive integers is 
smallest of these three integers. 



-66. Find the 



Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1 . Set up a Variable Dictionary. Let k represent the smallest of three con- 
secutive integers. 

2. Set up an Equation. An example of three consecutive integers is 34, 35, 
and 36. These are not the integers we seek, but they serve to help in the 
understanding of the problem. Note how each consecutive integer is one 
larger than the preceding integer. If k is the smallest of three consecutive 
integers, then the next two consecutive integers are k + 1 and k + 2. The 
"sum of three consecutive integers is —66" becomes: 



2.5. APPLICATIONS 



123 



First 




second 




third 


consecutive 


plus 


consecutive 


plus 


consecutive 


integer 




integer 




integer 



+ 



(fc + 1) 



(fc + 2) 



-66 



-66 



3. Solve the Equation. To solve for k, first simplify the left-hand side of the 
equation by combining like terms. 



fc + (fc + l) + (fc + 2) = 


-66 


Original equation. 


3fc + 3 = 


-66 


Combine like terms. 


3fc+3-3= 


-66-3 


Subtract 3 from both sides 


3fc = 


-69 


Simplify. 


3fc 

y ~ 


-69 
3 


Divide both sides by 3. 


k = 


-23 


Simplify. 



4. Answer the Question. The smallest of three consecutive integers is —23. 

5. Look Back. If k = — 23 is the smallest of three consecutive integers, then 
the next two consecutive integers are —22 and —21. Let's check the sum 
of these three consecutive integers. 



-23 +(-22) + (-21) 



-66 



Hence, the sum of the three consecutive integers is —66, as required. Our 

solution is correct. Answer: —77, —75, —73 



□ 



You Try It! 



EXAMPLE 3. A carpenter cuts a board measuring 60 inches in three pieces. 
The second piece is twice as long as the first piece, and the third piece is 
three times as long as the first piece. Find the length of each piece cut by the 
carpenter. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Let L represent the length of the first 
piece. Then the second piece, which is twice as long as the first piece, 
has length 2L. The third piece, which is three times as long as the first 
piece, has length 3L. Let's construct a little table to help summarize the 
information provided in this problem. 



Han cuts a board measuring 
230 inches in three pieces. 
The second piece is twice as 
long as the first piece, and 
the third piece is 30 inches 
longer than the second piece. 
Find the length of each piece 
cut by Han. 



124 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Piece 


Length (in) 


First piece 
Second piece 
Third piece 


L 

2L 
3L 


Total length 


60 



2. Set up an Equation. As you can see in the table above, the second column 
shows that the sum of the three pieces is 60 inches. In symbols: 

L + 2L + 3L = 60 



3. Solve the Equation. To solve for L, first simplify the left-hand side of the 
equation by combining like terms. 



L 



2L + 3L = 60 
6L = 60 
6L _ 60 

it ~ y 

L= 10 



Original equation. 
Combine like terms. 

Divide both sides by 6. 
Simplify. 



4. Answer the Question. The first piece has length L = 10 inches. The 
second piece has length 2L = 20 inches. The third piece has length 
3L = 30 inches. In tabular form, this is even more apparent. 



Piece 


Length (in) 


Length (in) 


First piece 
Second piece 
Third piece 


L 

2L 
3L 


10 

20 
30 


Total length 


60 


60 



Answer: 40, 80, 110 in 



5. Look Back. Not only is the second length twice the first and the third 
length three times the first, check the sum of their lengths: 

10 + 20 + 30 = 60 

That's a total of 60 inches. We have the correct solution. 



□ 



You Try It! 



The three sides of a triangle 
are consecutive integers. If 
the perimeter (sum of the 
three sides) of the triangle is 
453 centimeters, find the 
length of each side of the 
triangle. 



EXAMPLE 4. The three sides of a triangle are consecutive even integers. If 
the perimeter (sum of the three sides) of the triangle is 156 centimeters, find 
the length of each side of the triangle. 



2.5. APPLICATIONS 125 



Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. An example of three consecutive even 
integers is 18, 20, and 22. These are not the integers we seek, but they 
do give us some sense of the meaning of three consecutive even integers. 
Note that each consecutive even integer is two larger than the preceding 
integer. Thus, if k is the length of the first side of the triangle, then the 
next two sides are fc + 2 and fc+4. In this example, our variable dictionary 
will take the form of a well-labeled figure. 




k + 2 



2. Set up an Equation. The perimeter of the triangle is the sum of the three 
sides. If the perimeter is 156 centimeters, then: 

fc + (fc + 2) + (fc + 4) = 156 

3. Solve the Equation. To solve for k, first simplify the left-hand side of the 
equation by combining like terms. 



k + ( 


fc + 2) + (fc + 4) = 


■■ 156 


Original equation. 




3fc + 6 = 


: 156 


Combine like terms. 




3fc+6-6= 


156-6 


Subtract 6 from both sides 




3fc = 


: 150 


Simplify. 




3fc 

y ~ 


150 


Divide both sides by 3. 




k = 


50 


Simplify. 



4. Answer the Question. Thus, the first side has length 50 centimeters. 
Because the next two consecutive even integers are k + 2 = 52 and fc + 4 = 
54, the three sides of the triangle measure 50, 52, and 54 centimeters, 
respectively. 

5. Look Back. An image helps our understanding. The three sides are 
consecutive even integers. 



126 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



54 cm 




50 cm 



52 cm 



Note that the perimeter (sum of the three sides) is: 
50 cm + 52 cm + 54 cm = 156 cm 



(2.1) 



Answer: 150, 151, 152 cm 



Thus, the perimeter is 156 centimeters, as it should be. Our solution is 
correct. 



□ 



You Try It! 



The second angle of a 
triangle is three times bigger 
than the first angle. The 
third angle of the triangle is 
a 40 degrees larger than the 
second angle. How many 
degrees are in each angle? 



EXAMPLE 5. A well-known fact from geometry is the fact that the sum 
of the angles of a triangle is 180°. Suppose we have a triangle whose second 
angle is 10 degrees larger than twice its first angle and whose third angle is 
50 degrees larger than its first angle. Find the measure of each angle of the 
triangle. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. The Greek alphabet starts out with the 
letters a, j3, 7, 5, e, . . . , in much the same way that the English alphabet 
start out with the letters a, b, c, d, e, . . . . Mathematicians love to use 
Greek letters, especially in the study of trigonometry. The greek letter 9 
(pronounced "theta" ) is particularly favored in representing an angle of 
a triangle. So, we'll let 6 represent the degree measure of the first angle 
of the triangle. The second angle is 10 degrees larger than twice the first 
angle, so the second angle is 26 + 10. The third angle is 50 degrees larger 
than the first angle, so the third angle is 9 + 50. Again, we'll set up a 
well-labeled figure for our variable dictionary. 




2.5. APPLICATIONS 



127 



2. Set up an Equation. The sum of the angles is 180°, so: 

+ (20 + 10) + (0 + 50) = 180 

3. Solve the Equation. To solve for 0, first simplify the left-hand side of the 
equation by combining like terms. 

+ (20 + 10) + (0 + 50) = 180 Original equation. 

40 + 60 = 180 Combine like terms. 
40 + 60 - 60 = 180 - 60 Subtract 60 from both sides. 

40 = 120 Simplify. 
40 _ 120 
T ~~ ~T 

= 30 Simplify. 



Divide both sides by 4. 



4. Answer the Question. Thus, the first angle is = 30 degrees, the second 
angle is 20 + 10 = 70 degrees, and the third angle is + 50 = 80 degrees. 

5. Look Back. An image helps our understanding. Note that the second 
angle is 10 degrees larger than twice the first angle. Note that the third 
angle is 50 degrees larger than the first angle. 




Note that the sum of the angles is: 

30° + 70° + 80° = 180° (2.2) 

Thus, the sum of the three angles is 180 degrees, as it should be. We 
have the correct solution. 



Answer: 20°, 60°, 100° 



□ 



EXAMPLE 6. Martha inherits $21,000 and decides to invest the money 
in three separate accounts. The amount she invests in the second account is 
twice the amount she invests in the first account. The amount she invests in 



You Try It! 



Jim inherits $15,000. He 
invests part in a fund that 
pays 5% per year and the 
rest in a fund that pays 4% 
per year. At the end of one 
year, the combined interest 
from both investments was 
$4,250. How much did he 
invest in each fund? 



128 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

the third account is $1,000 more than the amount she invests in the second 
account. How much did she invest in each account? 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. We'll use a table in this example to help 
set up our variable dictionary. Let x be the amount invested in the 
first account. The amount invested in the second account is twice that 
invested in the first account, so 2x is the amount invested in the second 
account. The third account investment is $1,000 more than the amount 
invested in the second account, so that is 2x + 1000. 



Account # 


Amount Invested 


Account #1 
Account #2 
Account #3 


X 

2x 
2x+ 1000 


Total Invested 


21000 



2. Set up an Equation. The second column of the table reveals the required 
equation. The three investments must sum to $21,000. 

x + 2x+{2x + 1000) = 21000 



3. Solve the Equation. To solve for x, first simplify the left-hand side of the 
equation by combining like terms. 

x + 2x + (2x + 1000) = 21000 Original equation. 

5x + 1000 = 21000 Combine like terms. 

5x + 1000 - 1000 = 21000 - 1000 Subtract 1000 from both sides. 

5a; = 20000 Simplify. 
5x _ 20000 
~5 ~ 5 

x = 4000 Simplify. 



Divide both sides by 5. 



4. Answer the Question. Substitute x = 4000 in each entry of the second 
column of the table above to produce the results in the table below. 



Account # 



Amount Invested Amount Invested 



Account # 1 
Account #2 
Account^ 




1,000 
S,000 
),000 



Total Invested 



21000 



$21,000 



2.5. APPLICATIONS 



129 



Look Back. As we can see in our answer table, the amount $8,000 invested 
in the second account is twice the amount invested in the first account. 
The amount $9,000 invested in the third account is %1,000 more than the 
amount invested in the second account. Moreover, the total investment 
is: 

$4, 000 + $8, 000 + $9, 000 = $21, 000 (2.3) 

Thus, the total investment is $21,000, as it should be. We have the 
correct solution. 



Answer: $5,000 at 5% and 
$10,000 at 4%. 



□ 



You Try It! 



EXAMPLE 7. Jeff is hiking the 2,650-mile Pacific Crest Trail from Mexico 
to Canada. Shortly before he crosses over from Oregon into Washington he is 
four times as far from the beginning of the trail as he is from the end. How 
much further does he have to hike? 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Let d represent the distance left for Jeff 
to hike. Because Jeff is four times further from the beginning of the trail 
than the end, the distance Jeff already completed is Ad. Let's construct a 
little table to help summarize the information provided in this problem. 



Margaret is cycling along a 
lane that measures 100 
miles. If Magaret is four 
times as far from the start of 
the ride as she is from the 
finish, how many more miles 
does she have to go before 
she finishes her ride? 



Section of Trail 


Distance (mi) 


Distance to finish 
Distance from start 


d 

Ad 


Total distance 


2650 



2. Set up an Equation. As you can see in the table above, the second column 
shows that the sum of the two distances is 2650 miles. In symbols: 

d + 4d= 2650 



3. Solve the Equation. To solve for d, first simplify the left-hand side of the 
equation by combining like terms. 



d + Ad = 2650 
5d = 2650 
5d _ 2650 
~5~ ~~ 5 
d = 530 



Original equation. 
Combine like terms. 

Divide both sides by 5. 
Simplify. 



130 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



4. Answer the Question. Jeff still has 530 miles to hike. 

5. Look Back. Because the amount left to hike is d = 530 miles, Jeff's 
distance from the start of the trail is 4rf = 4(530), or 2,120 miles. If we 
arrange these results in tabular form, it is evident that not only is the 
distance from the start of the trail four times that of the distance left to 
the finish, but also the sum of their lengths is equal to the total length 
of the trail. 



Section of Trail 


Distance (mi) Distance (mi) 


Distance to finish 
Distance from start 


d 

Ad 


530 

2120 


Total distance 


2650 


2650 



Answer: 20 miles 



Thus, we have the correct solution. 



□ 



You Try It! 



20% of Mary's class were ill 
and stayed home from 
school. If only 36 students 
are present, what is the 
actual size of Mary's class? 



EXAMPLE 8. Today 15% of Sister Damaris' seventh grade class were ill and 
stayed home from school. If only 34 students are present, what is the actual 
size of Sister Damaris' class? 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 



1. Set up a Variable Dictionary. Let S represent the actual size of Sister 
Damaris' class. 

2. Set up an Equation. If 15% of Sister Damaris' class was absent, then 
85% of her class was present. There are 34 student present, so the phrase 
"85% of Sister Damaris' class is 34" translates into the equation, 

0.85,5 = 34, 

where we've changed 85% to a decimal by moving the decimal point two 
places to the left. 

3. Solve the Equation. To solve for S, first clear the decimals by multiplying 
both sides of the equation by 100. 



0.855 = 34 

85,5 = 3400 

85S* _ 3400 

"85" ~ 85 

S* = 40 



Original equation. 
Multiply both sides by 100. 

Divide both sides by 85. 
Simplify. 



2.5. APPLICATIONS 131 



4. Answer the Question. Sister Damaris' class size is 40. 

5. Look Back. We're told that 15% of Sister Damaris' class is absent. If we 
calculate 15% of 40, we get: 

0.15(40) = 6 

Thus, there were 6 students absent, so 40 — 6, or 34 students were present. 

Thus, we have the correct solution. Answer: 45 



□ 



132 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



f»- n- fa- 



Exercises 



■*1 ■*'» .*5 



1. Two angles are said to be complementary 
if their sum is ninety degrees. Suppose 
you have two complementary angles such 
that the second angle is 6 degrees larger 
than 2 times the measure of the first an- 
gle. Find the angles. 

2. Two angles are said to be complementary 
if their sum is ninety degrees. Suppose 
you have two complementary angles such 
that the second angle is 10 degrees larger 
than 3 times the measure of the first an- 
gle. Find the angles. 

3. Two angles are said to be supplementary 
if their sum is 180 degrees. Suppose you 
have two supplementary angles such that 
the second angle is 10 degrees larger than 

4 times the measure of the first angle. 
Find the angles. 

4. Two angles are said to be supplementary 
if their sum is 180 degrees. Suppose you 
have two supplementary angles such that 
the second angle is 12 degrees larger than 

5 times the measure of the first angle. 
Find the angles. 

5. The three sides of a triangle are consecu- 
tive integers. If the perimeter (sum of the 
three sides) of the triangle is 483 meters, 
find the length of each side of the triangle. 

6. The three sides of a triangle are consecu- 
tive integers. If the perimeter (sum of the 
three sides) of the triangle is 486 yards, 
find the length of each side of the trian- 
gle. 

7. Four less than eight times a certain num- 
ber is —660. Find the number. 

8. Nine less than five times a certain number 
is 141. Find the number. 



9. Alan is hiking a trail that is 70 miles long. 
After several days, he is four times as far 
from the beginning of the trail as he is 
from the end. How much further does he 
have to hike? 

10. Joe is hiking a trail that is 30 miles long. 
After several days, he is two times as far 
from the beginning of the trail as he is 
from the end. How much further does he 
have to hike? 

11. Martha takes roll in her sixth grade grade 
class and finds that 2 students are missing. 
If her actual class size is 36 students, what 
percentage of her class is absent? Round 
your answer to the nearest percent. 

12. Alice takes roll in her first grade grade 
class and finds that 7 students are missing. 
If her actual class size is 37 students, what 
percentage of her class is absent? Round 
your answer to the nearest percent. 

13. Lily cuts a piece of yarn into three pieces. 
The second pieces is 3 times as long as the 
first piece, and the third piece is 6 cen- 
timeters longer than the first piece. If the 
total length of the yarn is 211 centime- 
ters, find the lengths of each of the three 
pieces. 

14. Jane cuts a piece of twine into three pieces. 
The second pieces is 7 times as long as the 
first piece, and the third piece is 5 feet 
longer than the first piece. If the total 
length of the twine is 320 feet, find the 
lengths of each of the three pieces. 

15. The three sides of a triangle are consecu- 
tive even integers. If the perimeter (sum 
of the three sides) of the triangle is 450 
yards, find the length of each side of the 
triangle. 



2.5. APPLICATIONS 



133 



16. The three sides of a triangle are consecu- 
tive even integers. If the perimeter (sum 
of the three sides) of the triangle is 318 
feet, find the length of each side of the 
triangle. 

17. The perimeter of a triangle is 414 yards. 
The second side of the triangle is 7 times 
as long as the first side and the third side 
of the triangle is 9 yards longer than the 
first side. Find the lengths of each of the 
three sides of the triangle. 

18. The perimeter of a triangle is 54 inches. 
The second side of the triangle is 2 times 
as long as the first side and the third side 
of the triangle is 6 inches longer than the 
first side. Find the lengths of each of the 
three sides of the triangle. 

19. The sum of three consecutive odd integers 
is —543. Find the smallest of the three 
consecutive odd integers. 

20. The sum of three consecutive odd integers 
is —225. Find the smallest of the three 
consecutive odd integers. 

21. The sum of the angles of a triangle is 180°. 
In triangle AABC, the degree measure of 
angle B is 4 times the degree measure of 
angle A. The degree measure of angle C is 
30 degrees larger than the degree measure 
of angle A. Find the degree measures of 
each angle of triangle AABC. 

22. The sum of the angles of a triangle is 180°. 
In triangle AABC, the degree measure of 
angle B is 4 times the degree measure of 
angle A. The degree measure of angle C is 
60 degrees larger than the degree measure 
of angle A. Find the degree measures of 
each angle of triangle AABC. 

23. The sum of three consecutive integers is 

— 384. Find the largest of the three con- 
secutive integers. 

24. The sum of three consecutive integers is 

— 501. Find the largest of the three con- 
secutive integers. 

25. Seven more than two times a certain num- 
ber is 181. Find the number. 



26. Nine more than two times a certain num- 
ber is 137. Find the number. 

27. The three sides of a triangle are consecu- 
tive odd integers. If the perimeter (sum of 
the three sides) of the triangle is 537 feet, 
find the length of each side of the triangle. 

28. The three sides of a triangle are consecu- 
tive odd integers. If the perimeter (sum 
of the three sides) of the triangle is 471 
centimeters, find the length of each side 
of the triangle. 

29. A store advertises that it is offering a 14% 
discount on all articles purchased at the 
store. If Yao pays $670.80 for an article, 
what was the marked price for the article? 

30. A store advertises that it is offering a 12% 
discount on all articles purchased at the 
store. If Roberto pays $560.56 for an ar- 
ticle, what was the marked price for the 
article? 

31. The sum of three consecutive even integers 
is —486. Find the smallest of the three 
consecutive even integers. 

32. The sum of three consecutive even integers 
is —354. Find the smallest of the three 
consecutive even integers. 

33. Burt inherits $45,500. He decides to in- 
vest part of the inheritance in a mutual 
fund and the remaining part in a certifi- 
cate of deposit. If the amount invested 
in the certificate of deposit is $3,500 more 
than 6 times the amount invested in the 
mutual fund, find the amount invested in 
each account. 

34. Phoenix inherits $12,000. He decides to 
invest part of the inheritance in a mutual 
fund and the remaining part in a certifi- 
cate of deposit. If the amount invested 
in the certificate of deposit is $3,000 more 
than 8 times the amount invested in the 
mutual fund, find the amount invested in 
each account. 



134 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

$*■ j*- &*■ Answers •*$ *** •** 

1. 28, 62 degrees 19. -183 

3. 34, 146 degrees 21. 25°, 100°, 55° 

5. 160, 161, 162 meters 03 —127 

7. -82 



9. 14 miles 
11. 6% 

13. 41, 123, and 47 centimeters 
15. 148, 150, 152 yards 
17. 45, 315, and 54 yards 



25. 87 

27. 177, 179, 181 feet 

29. $780.00 

31. -164 

33. $6,000, $39,500 



2.6. INEQUALITIES 



135 



2.6 Inequalities 

In Chapter 1, we introduced the natural numbers N = {1, 2,3,.. .}, the whole 
numbers W = {0, 1, 2, 3, . . .}, and the integers Z = {. . . , -3, -2, -1, 0, 1, 2, 3, . . .}. 
Later in the chapter, we introduced the rational numbers, numbers of the form 
p/q, where p and q are integers. We noted that both terminating and repeating 
decimals are rational numbers. Each of these numbers has a unique position 
on the number line (see Figure 2.13). 



-3.125 



0.3 



1.5 



.'52 



-4-3-2-10 1 2 3 4 

Figure 2.13: Positioning numbers on the number line. 



The natural numbers, whole numbers, and integers are also rational num- 
bers, because each can be expressed in the form p/q, where p and q are integers. 
For example, 0=0/12, 4 = 4/1, and —3 = —12/4. Indeed, the rational numbers 
contain all of the numbers we've studied up to this point in the course. 

However, not all numbers are rational numbers. For example, consider the 
decimal number —3.10110111011110 . . ., which neither terminates nor repeats. 
The number v2 = 1.414213562373095 . . . also equals a decimal number that 
never terminates and never repeats. A similar statement can be made about 
the number 7r = 3.141592653589793 . . .. Each of these irrational (not rational) 
numbers also has a unique position on the number line (see Figure 2.14). 



-3.010110111.. 

—i w 1- 



V2 



7T 



-4-3-2-10 1 2 3 4 

Figure 2.14: Positioning numbers on the number line. 



The Real Numbers. If we combine all of the rational and irrational numbers 
into one collection, then we have a set of numbers that is called the set of real 
numbers. The set of real numbers is denoted by the symbol K. 



Two other irrational 
numbers you may encounter 
in your mathematical studies 
are e (Euler's constant), 
which is approximately equal 
to ew 2.71828182845904..., 
and (pronounced "phi" ) , 
called the golden ratio, which 
equals cp = (1 + v / 5)/2. The 
number e arises in 
applications involving 
compound interest, 
probability, and other areas 
of mathematics. The number 
4> is used in financial markets 
and is also arguably the ratio 
of beauty in art and 
architecture. 



Every point on the number line is associated with a unique real number. Con- 
versely, every real number is associated with a unique position on the number 
line. In lieu of this correspondence, the number line is usually called the real 
line. 



136 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Ordering the Real Numbers 

The real numbers are ordered on the real line in a manner identical to how we 
ordered the integers on the number line in Section 1 of Chapter 1 . 



Order on the Real Line. Suppose that a and b are real numbers positioned 
on the real line as shown below. 



• Because a lies to the "left of" b, we say that a is "less than" b, or in 
mathematical symbols, a < b. The inequality symbol < is read "less 
than." 

• Alternately, b lies to the "right of" a, so we can also say that b is "greater 
than" a, or in mathematical symbols, b > a. The inequality symbol > is 
read "greater than." 



Here are two more inequality symbols that we will use in this section. 

Less than or equal to. If we want to say that a lies to the "left of" 6, or 
shares the same position as 6, then we say that a is "less than or equal to" b 
and write a < b. The inequality symbol < is pronounced "less than or equal 
to." 



Greater than or equal to. If we want to say that b lise to the "right of" a, 
or shares the same position as a, then we say that b is "greater than ore equal 
to o and write b > a. The inequality symbol > is pronounced "greater than or 
equal to." 



Set-Builder Notation 

Mathematicians use a construct called set-builder notation to describe sets 
or collections of numbers. The general form of set-builder notation looks as 
follows: 

{x : some statement about x} 

For example, suppose that we want to describe the set of "all real numbers 
that are less than 2." We could use the following notation: 



{x : x < 2} 



2.6. INEQUALITIES 137 

This is read aloud as follows: U A equals the set of all x such that x is less than 
2." Some prefer to use a vertical bar instead of a colon. 

A = {x\x < 2} 

In this text we use the colon in set-builder notation, but feel free to use the 
vertical bar instead. They mean the same thing. 

One might still object that the notation {x : x < 2} is a bit vague. One 
objection could be "What type of numbers x are you referring to? Do you want 
the integers that are less than two or do you want the real numbers that are 
less than two?" As you can see, this is a valid objection. One way of addressing 
this objection is to write: 

A = {ieK:i<2) or A = {x e N : x < 2} 

The first is read "A is the set of all x in K that are less than two," while the 
second is read "A is the set of all x in N that are less than two." 

Set-builder Assumption. In this text, unless there is a specific refernce to 
the set of numbers desired, we will assume that the notation {x : x < 2} is 
asking for the set of all real numbers less than 2. 

In Figure 2.15, we've shaded the set of real numbers {x : x < 2}. Because 

^— I 1 1 1 1 1 1 O 1 1 1 — > 

-5-4-3-2-10 1 2 3 4 5 

Figure 2.15: Shading the numbers less than 2. 

"less than" is the same as saying "left of," we've shaded (in red) all points 
on the real line that lie to the left of the number two. Note that there is an 
"empty circle" at the number two. The point representing the number two is 
not shaded because we were only asked to shade the numbers that are strictly 
less than two. 

While the shading in Figure 2.15 is perfectly valid, much of the information 
provided in Figure 2.15 is unnecessary (and perhaps distracting). We only need 
to label the endpoint and shade the real numbers to the left of two, as we've 
done in constructing Figure 2.16. 

< 3 > 

2 

Figure 2.16: You only need to label the endpoint. 

For contrast, suppose instead that we're asked to shade the set of real 
numbers {x : x < 2}. This means that we must shade all the real numbers 



138 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Figure 2.17: You only need to label the endpoint. 



You Try It! 



Shade {x : x < 4} on the 
real line. 



Answer: 



that are "less than or equal to 2" or "left of and including 2." The resulting 
set is shaded in Figure 2.17. 

Note the difference between Figures 2.16 and 2.17. In Figures 2.16 we're 
shading the set {x : x < 2}, so the number 2 is left unshaded (an empty dot). 
In Figures 2.17, we're shading the set {x : x < 2}, so the number 2 is shaded 
(a filled- in dot). 



EXAMPLE 1. Shade the set {x : x > -3} on the real line. 

Solution: The notation {x : x > —3} is pronounced "the set of all real 
numbers x such that x is greater than or equal to —3." Thus, we need to shade 
the number —3 and all real numbers to the right of —3. 



<- 



-> 



□ 



You Try It! 



Use set-builder notation to 
describe the following set of 
real numbers: 



O- 
-10 



■¥ 



Answer: {x : x > —10} 



EXAMPLE 2. Use set-builder notation to describe the set of real numbers 
that are shaded on the number line below. 



«- 



O 



Solution: The number —1 is not shaded. Only the numbers to the left of —1 
are shaded. This is the set of all real numbers x such that x is "less than" —1. 
Thus, we describe this set with the following set-builder notation: 



{•-• 



x < 



■1} 



□ 



Interval Notation 

In Examples 1 and 2, we used set-builder notation to describe the set of real 
numbers greater than or equal to —3 and a second set of real numbers less than 
— 1. There is another mathematical symbolism, called interval notation, that 
can be used to describe these sets of real numbers. 

Consider the first set of numbers from Example 1, {x : X > —3}. 



2.6. INEQUALITIES 139 



Sweeping our eyes "from left to right", we use [— 3,oo) to describe this set of 
real numbers. Some comments are in order: 

1. The bracket at the left end means that —3 is included in the set. 

2. As you move toward the right end of the real line, the numbers grow 
without bound. Hence, the oo symbol (positive infinity) is used to indi- 
cate that we are including all real numbers to the right of —3. However, 
oo is not really a number, so we use a parentheses to indicate we are "not 
including" this fictional point. 

The set of numbers from Example 2 is {a: : x < —I}. 



< o > 

-1 

Sweeping our eyes "from left to right" , this set of real numbers is described 
with (— oo, —1). Again, comments are in order: 

1. The number —1 is not included in this set. To indicate that it is not 
included, we use a parenthesis. 

2. As you move toward the left end of the real line, the numbers decrease 
without bound. Hence, the — oo symbol (negative infinity) is used to 
indicate that we are including all real numbers to the left of —1. Again, 
— oo is not an actual number, so we use a parenthesis to indicate that we 
are not including this "fictional" point. 



Sweep your eyes from "left to right" . If you would like to insure that you 
correctly use interval notation, place the numbers in your interval notation in 
the same order as they are encountered as you sweep your eyes from "left to 
right" on the real line. 



A nice summary of set-builder and interval notation is presented in Table 2.1 
at the end of the section. 



Equivalent Inequalities 

Like equations, two inequalities are equivalent if they have the same solution 

sets. 



140 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Adding or Subtracting the Same Quantity from Both Sides of an 
Inequality. Let a and b be real numbers with 



If c is any real number, then 



and 



a < b, 



a + c < b + c 



c < b — c. 



That is, adding or subtracting the same amount from both sides of an inequality 
produces an equivalent inequality (does not change the solution). 



You Try It! 



Use interval notation to 
describe the solution of: 

x-7 < -8 



Answer: 



-oo,—l] 



EXAMPLE 3. Solve for a;: x - 2 < 7. Sketch the solution on the real line, 
then use set-builder and interval notation to describe your solution. 

Solution: To "undo" subtracting 2, we add 2 to both sides of the inequality. 

x — 2 < 7 Original inequality. 

s-2 + 2<7 + 2 Add 2 to both sides. 

x < 9 Simplify both sides. 

To shade the real numbers less than or equal to 9, we shade the number 9 and 
all real numbers to the left of 9. 

< • > 



9 

Using set-builder notation, the solution is {x : x < 9}. Using interval notation, 
the solution is (— oo,9]. 

□ 

If we multiply or divide both sides of an inequality by a positive number, 
we have an equivalent inequality. 



Multiplying 


or 


Dividing 


by 


a Positive 


Number. 


Let 


a 


and b be 


real 


numbers with 


a < 


b. If c is a 


rea] 


positive number, 


then 


















ac < be 














and 








a b 
- < -. 

c c 















2.6. INEQUALITIES 



141 



EXAMPLE 4. Solve for x: 3x < —9 Sketch the solution on the real line, 
then use set-builder and interval notation to describe your solution. 

Solution: To "undo" multiplying by 3, divide both sides of the inequality by 
3. Because we are dividing both sides by a positive number, we do not reverse 
the inequality sign. 



You Try It! 



Use interval notation to 
describe the solution of: 

2x > -8 



3x < 


-9 


Original inequality. 


3ir 
— < 

3 ~ 


-9 
~3~ 


Divide both sides by 3 


x < 


-3 


Simplify both sides. 



Shade the real numbers less than or equal to —3. 



Using set-builder notation, the solution is {x : x < —3}. Using interval nota- 
tion, the solution is (— oo, —3]. 



Answer: (—4, oo) 



Reversing the Inequality Sign 

Up to this point, it seems that the technique for solving inequalities is pretty 
much identical to the technique used to solve equations. However, in this 
section we are going to encounter one exception. 

Suppose we start with the valid inequality — 2 < 5, then we multiply both 
sides by 2, 3, and 4. 



□ 



-2 < 5 


-2 < 5 


-2 < 5 


2(-2) < 2(5) 


3(-2) <3(5) 


4(-2) < 4(5) 


-4< 10 


-6 < 15 


-8 < 20 



Note that in each case, the resulting inequality is still valid. 



Caution! We're about to make an error! Start 


again with — 2 < 5, but 


this time multiply both sides by —2, —3, and —4. 




-2 < 5 -2 < 5 


-2 < 5 


-2(-2) < -2(5) -3(-2)<-3(5) 


-4(-2) < -4(5) 


4<-10 6<-15 


8< -20 



142 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



Note that in each of the resulting inequalities, the inequality symbol is pointing 
the wrong way! 



Some readers might prefer a 
more formal reason as to 
why we reverse the 
inequality when we multiply 
both sides by a negative 
number. Suppose that a < b. 
Then, subtracting b from 
both sides gives the result 
a — b < 0. This means that 
a — b is a negative number. 
Now, if c is a negative 
number, then the product 
(a — b)c is positive. Then: 

(a-b)c> 
ac — be > 
ac — be + be > + be 
ac > be 

Thus, if you start with a < b 
and c < 0, then ac > be. 



You Try It! 



Use interval notation to 
describe the solution of: 

-3a; > -6 



When you multiply both sides of an inequality by a negative number, you must 
reverse the inequality sign. Starting with — 2 < 5, multiply both sides by —2, 
—3, and —4, but reverse the inequality symbol. 



-2 < 5 
-2(-2) > -2(5) 
4> -10 



-2 < 5 
-3(-2) > -3(5) 
6 > -15 



-2 < 5 

-4(-2) > -4(5) 
8 > -20 



Note that each of the resulting inequalities is now a valid inquality. 



Multiplying or Dividing by a Negative Number. 


Let 


a and b be real 


numbers 


i with a < b. If c is 


a real negative number, 
ac > be 


then 






and 




a b 
- > -. 

c c 








That is, 


when multiplying or dividing both sides of 


an inequality by a negative 


number 


you must reverse 


the inequality sign. 









EXAMPLE 5. Solve for x: — 2x < 4. Sketch the solution on the real line, 
then use set-builder and interval notation to describe your solution. 

Solution: To "undo" multiplying by —2, divide both sides by —2. Because we 
are dividing both sides by a negative number, we reverse the inequality sign. 



-2x <4 
-2x 4 
^2 > ~2 



x > 



Original inequality. 

Divide both sides by —2. 
Reverse the inequality sign. 

Simplify both sides. 



Shade the real numbers greater than —2. 



-O 



Answer: (— oo,2] 



Using set-builder notation, the solution is {x : x > — 2}. Using interval nota- 
tion, the solution is (— 2,oo). 



□ 



2.6. INEQUALITIES 



143 



Multiple Steps 

Sometimes you need to perform a sequence of steps to arrive at the solution. 



EXAMPLE 6. Solve for x: 2x + 5 > -7. Sketch the solution on the real 
line, then use set-builder and interval notation to describe your solution. 

Solution: To "undo" adding 5, subtract 5 from both sides of the inequality. 

2x + 5 > — 7 Original inequality. 

2a; + 5 — 5>— 7— 5 Subtract 5 from both sides. 

2a; > — 12 Simplify both sides. 

To "undo" multiplying by 2, divide both sides by 2. Because we are dividing 
both sides by a positive number, we do not reverse the inequality sign. 



2.r 



-12 



> 



x > 



Divide both sides by 2. 
Simplify both sides. 



Shade the real numbers greater than —6. 



-O- 
-6 



You Try It! 



Use interval notation to 
describe the solution of: 

3a; - 2 < 4 



Using set-builder notation, the solution is {x : x > —6}. Using interval nota- 
tion, the solution is (—6, oo). Answer: (— oo,2] 



□ 



EXAMPLE 7. Solve for x: 3 - 5a: < 2a: + 17. Sketch the solution on the 
real line, then use set-builder and interval notation to describe your solution. 

Solution: We need to isolate terms containing x on one side of the inequality. 
Start by subtracting 2a; from both sides of the inequality. 



3- 5a; < 2a; + 17 
5a; - 2a: < 2a; + 17 - 
3 -7a; < 17 



Original inequality. 

2a; Subtract 2a; from both sides. 

Simplify both sides. 



We continue to isolate terms containing x on one side of the inequality. Subtract 
3 from both sides. 



3- 7a;- 3 < 17 
-7a; < 14 



Subtract 3 from both sides. 
Simplify both sides. 



You Try It! 



Use interval notation to 
describe the solution of: 

4 - x > 2x + 1 



144 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

To "undo" multiplying by —7, divide both sides by — 7. Because we are dividing 
both sides by a negative number, we reverse the inequality sign. 



-Ix 14 
— > — 

-7 " -7 



x > -2 
Shade the real numbers greater than or equal to —2 



Divide both sides by —7. 
Simplify both sides. 



Answer: (—00, 1) 



You Try It! 



Use interval notation to 
describe the solution of: 



2x 

T 



> 



Using set-builder notation, the solution is {x : x > — 2}. Using interval nota- 
tion, the solution is [— 2,oo). 

□ 

We clear fractions from an inequality in the usual manner, by multiplying 
both sides by the least common denominator. 



EXAMPLE 8. Solve for 



x_ 1 

12 > 3' 



Solution: First, clear the fractions from the inequality by multiplying both 
sides by the least common denominator, which in this case is 12. 



x 1 
12 > 3 



12 


3 x 

4 ~ 12 


> 


'1" 
3 


"3' 

4_ 




1.12- 


> 


T 
3 



9- a; >4 



Original inequality. 
12 Multiply both sides by 12. 

12 Distribute the 12. 

Cancel and multiply. 



To "undo" adding 9, subtract 9 from both sides. 



9-2-9 >4-9 
-x > -5 



Subtract 9 from both sides. 
Simplify both sides. 



We could divide both sides by —1, but multiplying both sides by —1 will also 
do the job. Because we are multiplying both sides by a negative number, we 
reverse the inequality sign. 



(-l)(-x) < (-5)(-l) 
x < 5 



Multiply both sides by —1. 
Reverse the inequality sign. 

Simplify both sides. 



2.6. INEQUALITIES 



145 



Shade the real numbers less than 5. 
i 



-O- 

5 



Using set-builder notation, the solution is {x : x < 5}. Using interval notation, 

the solution is (— oo, 5). Answer: [— 9/8, oo) 



We clear decimals from an inequality in the usual manner, by multiplying 
both sides by the appropriate power of ten. 



□ 



EXAMPLE 9. Solve for x: 3.25 - 1.2a; > 4.6. 

Solution: First, clear the decimals from the inequality by multiplying both 
sides by 100, which moves each decimal point two places to the right. 



3.25- 1.2cc > 4.6 
325 - 120a; > 460 
325 - 120a; - 325 > 460 - 325 
-120a; > 135 

-120a; 135 

< 



-120 



x < 



-120 

27 
24 



Original inequality. 
Multiply both sides by 100. 
Subtract 325 from both sides. 
Simplify both sides. 

Divide both sides by —120. 
Reverse the inequality sign. 

Reduce to lowest terms. 



Shade the real numbers less than —27/24. 



«- 



O 



-27/24 

Using set-builder notation, the solution is {x : x < —27/24}. Using interval 
notation, the solution is (-co, —27/24). 



You Try It! 



Use interval notation to 
describe the solution of: 

2.3a; -5.62 > -1.4 



Answer: [211/115, oo) 



□ 



146 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

Summary Table of Set-Builder and Interval Notation 

A summary table of the set-builder and interval notation is presented in Table 2.1. 



Shading on the real I 


ine 


Set-builder 


Interval 






{x : x > —5} 




-5 




(-5,oo) 




_^ 


{x : x > —5} 




-5 


^ 


[-5,oo) 




— > 


{x : x < —5} 




-5 


(-oo,-5) 


/ A 




{x : x < —5} 




-5 




(-oo,-5] 



Table 2.1: Examples of set-builder and interval notation. 



2.6. INEQUALITIES 



147 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



1. Draw a number line, then plot the num- 
bers 4, 3, —4, 7/8, and —8/3 on your num- 
ber line. Label each point with its value. 
Finally, list the numbers in order, from 
smallest to largest. 

2. Draw a number line, then plot the num- 
bers 5, 3, —4, 5/7, and —4/3 on your num- 
ber line. Label each point with its value. 
Finally, list the numbers in order, from 
smallest to largest. 



Draw a number line, then plot the num- 
bers — 5, 5, 4, 2/3, and 8/3 on your num- 
ber line. Label each point with its value. 
Finally, list the numbers in order, from 
smallest to largest. 

Draw a number line, then plot the num- 
bers — 3, —2, 4, 1/3, and 5/2 on your num- 
ber line. Label each point with its value. 
Finally, list the numbers in order, from 
smallest to largest. 



In Exercises 5-20, shade each of the following sets on a number line. 



5. {x : x > -7} 

6. {x : x > -1} 

7. {x : x < 2} 

8. {x : x < -6} 

9. (-oo,2) 

10. (-oo,-9) 

11. (6,00) 

12. (5,oo) 



13. {x : x > 7} 

14. {x : x > -8} 

15. [0,oo) 

16. [7,oo) 

17. {x : x < -2} 

18. {x : x < 7} 

19. (-oo,3] 

20. (-oo,-l] 



In Exercises 21-28, use set-builder notation to describe the shaded region on the given number line. 



21. <- 



22. «- 



23. <- 



24. <- 



9 

-•- 


■O 



-o- 

9 



25. 
26. 

27. 
28. 



-O 



-o 



148 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 

In Exercises 29-36, use interval notation to describe the shaded region on the given number line. 



29. < O > 33. «- 

4 



30. < O > 34. «- 

-7 

31. < O > 35. <- 



32. < O > 36. 

2 



In Exercises 37-44, solve each of the given inequalities. Sketch the solution on a number line, then use 
set-builder and interval notation to describe your solution. 

37. a; + 10< 19 41. -2x < -2 

38. x + 17 > 7 42. -18a; > -20 

39. Ax < 8 43. x- 18 > -10 

40. 16a; > -2 44. x - 8 < -18 



In Exercises 45-62, solve each of the given inequalities. Sketch the solution on a number line, then use 
set-builder and interval notation to describe your solution. 

45. -5a; -6 > 4 - 9a; 54. -11a; - 7 > -15 - 5x 

46. 2a;- 7 > -3- 4a; 55. 2a; - 9 > 5 - 8a; 

47. 16a;- 6 < 18 56. -3a; - 6 > -2 - 9a; 

48. 8a;- 14 < -12 57. -10a; - 4 < 18 

49. -14a;- 6 > -10- 4a; 58. -6a;- 14 <1 

50. -13a;- 4 > -2- 5a; 59. -12a; + 4 < -56 

51. 5a; + 18 < 38 60. -18a; + 6 < -12 

52. 9a; + 16 < 79 61. 15a; + 5 < 6a; + 2 

53. -16a;- 5 > -11 -6a; 62. 12a; + 8 < 3a; + 5 



2.6. INEQUALITIES 



149 



In Exercises 63-76, solve each of the given inequalities. Sketch the solution on a number line, then use 
set-builder and interval notation describe your solution. 



63. 


3 9 
-x > - 

2 8 




64. 


6 3 

7 X > 7 

7 4 




65. 


3 9 

X+ 2<5 




66. 


1 1 

x +l < ~7 
4 5 




67. 


4 1 4 

x < —x - 

7 6~3 


1 
" 2 


68. 


5 3 7 

x < —x - 

3 4-4 


3 

" 5 


69. 


3 9 
x > — 

8 " 7 





70. 


7 1 

a? > - 

2-5 


71. 


6 4 
5^-7 


72. 


4 2 
-x < - 
3-9 


73. 


6 7 5 2 

— x < X 

5 3-99 


74. 


3 13 2 
— x < X 

7 2~2 7 


75. 


9 9 17 

—x H — > —x H — 

7 2 7 2 


76. 


5 9 15 
—x H — > —x H — 
7 2 3 2 



In Exercises 77-84, solve each of the given inequalities. Sketch the solution on a number line, then use 
set-builder and interval notation containing fractions in reduced form to describe your solution. 



77. -3.7a;- 1.98 < 3.2 

78. -3.6a; -3.32 < 0.8 

79. -3.4a; + 3.5 > 0.9 - 2.2.x 

80. -2.6a; + 3.1 > -2.9- 1.7x 



81. -1.3a; + 2.9 > -2.6 - 3.3x 

82. 2.5x + 2.1 > 1.4 -3.8a; 

83. 2.2x + 1.9 < -2.3 

84. 1.6a: +1.2 < 1.6 



j*- j*- f*- Answers •*$ ■*$ ■** 



3. 
5. 



3 4 



-5-4-3-2-10 1 2 3 4 5 



3 4 5 

— •-! • • > 



-5-4-3-2-10 1 2 3 4 5 



9. <- 

11. <- 

13. <- 
15. <- 



O 
2 



-O 
6 

-O 

7 



7. <- 



O 

2 



17. <- 



150 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 



19. <- 



21. 


{x : x < 9} 


23. 


{x : x < -8} 


25. 


{x : x > -2} 


27. 


{x : x > —3} 


29. 


(4,oo) 


31. 


(-oo,-2) 


33. 


(— oo, 5] 


35. 


[l,oo) 


37. 


(-oo,9) 


39. 


(-oo,2) 


41. 


[l,oo) 


43. 


(8,00) 


45. 


[5/2,oo) 


47. 


(-oo,3/2] 


49. 


[2/5,oo) 


51. 


(-oo,4) 



53. 


3/5, 00) 


55. 


7/5,oo) 


57. 


-ll/5,oo) 


59. 


5, 00) 


61. 


-oo,-l/3) 


63. 


3/4,oo) 


65. 


-oo,3/10) 


67. 


5/7,oo) 


69. 


-51/56, 00) 


71. 


10/21,oo) 


73. 


-65/22, 00) 


75. 


-7/8,oo) 


77. 


-7/5,oo) 


79. 


-oo,13/6] 


81. 


-ll/4,oo) 


83. 


-oo,-21/ll 



Chapter 3 



Introduction to Graphing 



In thinking about solving algebra and geometry problems in the 1600's, Rene 
Descartes came up with the idea of using algebraic procedures to help solve 
geometric problems and to also demonstrate algebraic operations by using ge- 
ometry. 

One of the easy ways to do this was to use the 2-dimensional geometric 
plane. He decided to use the letters near the end of the alphabet to name the 
axes: "x-axis" for the horizontal axis and "y-axis" for the vertical axis. Then 
ordered pairs (x, y) would be used to graph specific points on the plane and/or 
algebraic expressions and equations. Thus, the graphs on the following pages 
are done on a graph named a "Cartesian Coordinate" graph in Rene Descartes' 
honor. 

The equation of a particular line can be written in several algebraic forms 
which you will learn to interpret. The geometric aspect of a line will be noted 
in its position and slope. We will also be focusing on geometric aspects such as 
parallel and perpendicular lines. Ultimately, these tasks will be used to help 
solve application problems involving speed and distance. 

In modern times, we use the graphing calculator to speed up the process 
of providing tables of data, graphing lines quickly, and finding solutions of 
equations. 



151 



152 



CHAPTER 3. INTRODUCTION TO GRAPHING 



3.1 Graphing Equations by Hand 

We begin with the definition of an ordered pair. 

Ordered Pair. The construct (x,y), where x and y are any real numbers, is 
called an ordered pair of real numbers. 

(4,3), (—3,4), (—2,-3), and (3,-1) are examples of ordered pairs. 

Order Matters. Pay particular attention to the phrase "ordered pairs." Or- 
der matters. Consequently, the ordered pair (x, y) is not the same as the 
ordered pair (y,x), because the numbers are presented in a different order. 



Rene Descartes (1596-1650) 
was a French philosopher 
and mathematician who is 
well known for the famous 
phrase "cogito ergo sum" (I 
think, therefore I am), which 
appears in his Discours de la 
methode pour bien conduire 
sa raison, et chercher la 
verite dans les sciences 
(Discourse on the Method of 
Rightly Conducting the 
Reason, and Seeking Truth 
in the Sciences). In that 
same treatise, Descartes 
introduces his coordinate 
system, a method for 
representing points in the 
plane via pairs of real 
numbers. Indeed, the 
Cartesian plane of modern 
day is so named in honor of 
Rene Descartes, who some 
call the "Father of Modern 
Mathematics" 



The Cartesian Coordinate System 

Pictured in Figure 3.1 is a Cartesian Coordinate System. On a grid, we've 
created two real lines, one horizontal labeled x (we'll refer to this one as the 
x-axis), and the other vertical labeled y (we'll refer to this one as the y-axis). 



y 



-5 -4 -3 -2 -1 



->■ x 



12 3 4 5 



-1 

-2 

-3 

-4- 

-5- 

Figure 3.1: The Cartesian Coordinate System. 



Two Important Points: Here are two important points to be made about 
the horizontal and vertical axes in Figure 3.1. 



3. 1 . GRAPHING EQUATIONS BY HAND 



153 



1. As you move from left to right along the horizontal axis (the x-axis in 
Figure 3.1), the numbers grow larger. The positive direction is to the 
right, the negative direction is to the left. 

2. As you move from bottom to top along the vertical axis (the y-axis in 
Figure 3.1), the numbers grow larger. The positive direction is upward, 
the negative direction is downward. 

Additional Comments: Two additional comments are in order: 

3. The point where the horizontal and vertical axes intersect in Figure 3.2 
is called the origin of the coordinate system. The origin has coordinates 
(0,0). 

4. The horizontal and vertical axes divide the plane into four quadrants, 
numbered I, II, III, and IV (roman numerals for one, two, three, and 
four), as shown in Figure 3.2. Note that the quadrants are numbered in 
a counter-clockwise order. 



II 



/ 



Origin: (0,0) 



5 - 


4 - 


3 - 


2 - 












Tl 


"T 






1J 


.1 













IV 



Figure 3.2: Numbering the quadrants and indicating the coordinates of the 
origin. 



154 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Plotting Ordered Pairs 

Before we can plot any points or draw any graphs, we first need to set up a 
Cartesian Coordinate System on a sheet of graph paper? How do we do this? 
What is required? 



We don't always label the 
horizontal axis as the x-axis 
and the vertical axis as the 
y-axis. For example, if we 
want to plot the velocity of 
an object as a function of 
time, then we would be 
plotting points (t, v). In that 
case, we would label the 
horizontal axis as the t-axis 
and the vertical axis as the 
u-axis. 



Draw and label each axis. If we are going to plot points (x,y), then, on a 
sheet of graph paper, perform each of the following initial tasks. 

1. Use a ruler to draw the horizontal and vertical axes. 

2. Label the horizontal axis as the x-axis and the vertical axis as the y-axis. 

The result of this first step is shown in Figure 3.3. 

Indicate the scale on each axis. 

3. Label at least one vertical gridline with its numerical value. 

4. Label at least one horizontal gridline with its numerical value. 



The scales on the horizontal 
and vertical axes may differ. 
However, on each axis, the 
scale must remain consistent. 
That is, as you count to the 
right from the origin on the 
x-axis, if each gridline 
represents one unit, then as 
you count to the left from 
the origin on the x-axis, each 
gridline must also represent 
one unit. Similar comments 
are in order for the y-axis, 
where the scale must also be 
consistent, whether you are 
counting up or down. 



An example is shown in Figure 3.4. Note that the scale indicated on the x-axis 
indicates that each gridline counts as 1-unit as we count from left-to-right. The 
scale on the y-axis indicates that each gridlines counts as 2-units as we count 
from bottom-to-top. 



.'/ 



■*■ x 



-10 



-10- 



* x 



Figure 3.3: Draw and label each 
axis. 



Figure 3.4: Indicate the scale on 
each axis. 



3. 1 . GRAPHING EQUATIONS BY HAND 



155 



Now that we know how to set up a Cartesian Coordinate System on a sheet 
of graph paper, here are two examples of how we plot points on our coordinate 
system. 



To plot the ordered pair (4, 3), start 
at the origin and move 4 units to 
the right along the horizontal axis, 
then 3 units upward in the direction 
of the vertical axis. 



(4 3) 



To plot the ordered pair (—2,-3), 
start at the origin and move 2 units 
to the left along the horizontal axis, 
then 3 units downward in the direc- 
tion of the vertical axis. 

y 



(-2,-3) 



Continuing in this manner, each ordered pair (x, y) of real numbers is asso- 
ciated with a unique point in the Cartesian plane. Vice-versa, each point in 
the Cartesian point is associated with a unique ordered pair of real numbers. 
Because of this association, we begin to use the words "point" and "ordered 
pair" as equivalent expressions, sometimes referring to the "point" (x, y) and 
other times to the "ordered pair" (x,y). 



EXAMPLE 1. Identify the coordinates of the point P in Figure 3.5. 

Solution: In Figure 3.6, start at the origin, move 3 units to the left and 4 
units up to reach the point P. This indicates that the coordinates of the point 
P are (-3,4). 



You Try It! 



Identify the coordinates of 
the point P in the graph 
below. 

y 











V 




























































































5 


















c. 




































/ 


) 































5 


■ 











156 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Answer: (3,-2) 



P{=3^) 



->■ x 



*■ x 



Figure 3.5: Identify the coordinates 
of the point P. 



Figure 3.6: Start at the origin, 
move 3 units left and 4 units up. 



□ 



The variables do not have to 
always be x and y. For 
example, the equation 
v = 2 + 3.2£ is an equation in 
two variables, v and t. 



Equations in Two Variables 

The equation y = x + 1 is an equation in two variables, in this case x and y. 
Consider the point (x, y) = (2,3). If we substitute 2 for x and 3 for y in the 
equation y = x + 1, we get the following result: 



x 
2- 

3 



Original equation. 
Substitute: 2 for x, 3 for y. 
Simplify both sides. 



Because the last line is a true statement, we say that (2, 3) is a solution of 
the equation y = x + 1. Alternately, we say that (2,3) satisfies the equation 
y = x + 1. 

On the other hand, consider the point (x, y) = (—3, 1). If we substitute —3 
for x and 1 for y in the equation y = x + 1, we get the following result. 



y 
l 

1 : 



x + 1 
-3 + 1 
-2 



Original equation. 
Substitute: —3 for x, 1 for y. 
Simplify both sides. 



Because the last line is a false statement, the point (—3, 1) is not a solution of 
the equation y = x+1; that is, the point (—3, 1) does not satisfy the equation 
y = x + 1. 



3. 1 . GRAPHING EQUATIONS BY HAND 



157 



Solutions of an equation in two variables. Given an equation in the 
variables x and y and a point (x, y) = (a, b), if upon subsituting a for x and b for 
y a true statement results, then the point (x, y) = (a, b) is said to be a solution 
of the given equation. Alternately, we say that the point (x, y) = (a, b) satisfies 
the given equation. 



You Try It! 



EXAMPLE 2. Which of the ordered pairs (0,-3) and (1,1) satisfy the Which of the ordered pairs 



equation y = 3x — 2? 

Solution: Substituting the ordered pairs (0, —3) and (1, 1) into the equation 
y = 3x — 2 lead to the following results: 



(—1,3) and (2, 1) satisfy the 
equation y = 2x + 5? 



Consider (x,y) = (0, —3). 
Substitute for x and —3 for y: 

y = 3 a: — 2 
-3 = 3(0) -2 
-3 = -2 

The resulting statement is false. 



Consider (x, y) = (1, 1). 
Substitute 1 for x and 1 for y: 

y = 3a; — 2 
1 = 3(1) -2 
1 = 1 

The resulting statement is true. 

3a; — 2, but 



Thus, the ordered pair (0, —3) does not satisfy the equation y 
the ordered pair (1, 1) does satisfy the equation y = 3a; — 2. 



Answer: (—1,3) 



Graphing Equations in Two Variables 

Let's first define what is meant by the graph of an equation in two variables. 

The graph of an equation. The graph of an equation is the set of all points 
that satisfy the given equation. 



□ 



EXAMPLE 3. Sketch the graph of the equation y 



1. 



Solution: The definition requires that we plot all points in the Cartesian 
Coordinate System that satisfy the equation y = x + 1. Let's first create a 
table of points that satisfy the equation. Start by creating three columns with 
headers x, y, and (a;, y), then select some values for x and put them in the first 
column. 



You Try It! 



Sketch the graph of the 
equation y = —x + 2. 



158 



CHAPTER 3. INTRODUCTION TO GRAPHING 



X 


y -■ 


= x + l 


(x,y) 


-3 




-2 


(-3,-2) 


-2 








-1 

















1 








2 








3 









Take the first value of x, namely x = 
—3, and substitute it into the equation 
y = x + 1. 

y = x + 1 
y = -3 + i 

2/ = -2 

Thus, when x = —3, we have y = —2. 
Enter this value into the table. 



Continue substituting each tabular value of x into the equation y = x + 1 and 
use each result to complete the corresponding entries in the table. 



y = -3 + 1 = -2 

2/=-2 + l = -l 
y = -1 + 1 = 

y = 0+1 = 1 

y=l+l=2 

y = 2 + 1 = 3 
y = 3+1 =4 



The last column of the table now contains seven points that satisfy the equation 
y = x+l. Plot these points on a Cartesian Coordinate System (see Figure 3.7). 



X 


y = x + 1 


(&>?/) 


-3 


-2 


(-3,-2) 


-2 


-1 


(-2,-1) 


-1 





(-1,0) 





1 


(0,1) 


1 


2 


(1,2) 


2 


3 


(2,3) 


3 


4 


(3,4) 



y 



v 



Figure 3.7: Seven points that sat- 
isfy the equation y = x + 1. 



Figure 3.8: Adding additional 
points to the graph of y = X + 1. 



In Figure 3.7, we have plotted seven points that satisfy the given equation 
y = x + 1. However, the definition requires that we plot all points that satisfy 
the equation. It appears that a pattern is developing in Figure 3.7, but let's 
calculate and plot a few more points in order to be sure. Add the a;-values 



3. 1 . GRAPHING EQUATIONS BY HAND 



159 



— 2.5, —1.5, —0.5, 05, 1.5, and 2.5 to the x-column of the table, then use the 
equation y = x + 1 to evaluate y at each one of these x-values. 



y = -2.5 + 1 = -1.5 
y = -1.5 + 1 = -0.5 
y = -0.5 + 1 =0.5 
y = 0.5 + 1 = 1.5 
y = 1.5 + 1 = 2.5 
y = 2.5 + 1 = 3.5 



X 


y = x + 1 


(x, y) 


-2.5 


-1.5 


(-2.5,-1.5) 


-1.5 


-0.5 


(-1.5,-0.5) 


-0.5 


0.5 


(-0.5,0.5) 


0.5 


1.5 


(0.5,1.5) 


1.5 


2.5 


(1.5,2.5) 


2.5 


3.5 


(2.5,3.5) 



Add these additional points to the graph in Figure 3.7 to produce the image 
shown in Figure 3.8. 

There are an infinite number of points that satisfy the equation y = x + 1. 
In Figure 3.8, we've plotted only 13 points that satisfy the equation. However, 
the collection of points plotted in Figure 3.8 suggest that if we were to plot the 
remainder of the points that satisfy the equation y = x + 1, we would get the 
graph of the line shown in Figure 3.9. 




Figure 3.9: The graph of y = x + 1 is a line. 




□ 



Guidelines and Requirements 

Example 3 suggests that we should use the following guidelines when sketching 
the graph of an equation. 



160 CHAPTER 3. INTRODUCTION TO GRAPHING 



Guidelines for drawing the graph of an equation. When asked to draw 
the graph of an equation, perform each of the following steps: 

1. Set up and calculate a table of points that satisfy the given equation. 

2. Set up a Cartesian Coordinate System on graph paper and plot the points 
in your table on the system. Label each axis (usually x and y) and 
indicate the scale on each axis. 

3. If the number of points plotted are enough to envision what the shape of 
the final curve will be, then draw the remaining points that satisfy the 
equation as imagined. Use a ruler if you believe the graph is a line. If 
the graph appears to be a curve, freehand the graph without the use of 
a ruler. 

4. If the number of plotted points do not provide enough evidence to en- 
vision the final shape of the graph, add more points to your table, plot 
them, and try again to envision the final shape of the graph. If you still 
cannot predict the eventual shape of the graph, keep adding points to 
your table and plotting them until you are convinced of the final shape 
of the graph. 



Here are some additional requirements that must be followed when sketch- 
ing the graph of an equation. 

Graph paper, lines, curves, and rulers. The following are requirements 
for this class: 

5. All graphs are to be drawn on graph paper. 

6. All lines are to be drawn with a ruler. This includes the horizontal and 
vertical axes. 

7. If the graph of an equation is a curve instead of a line, then the graph 
should be drawn freehand, without the aid of a ruler. 



Using the TABLE Feature of the Graphing Calculator 

As the equations become more complicated, it can become quite tedious to 
create tables of points that satisfy the equation. Fortunately, the graphing 
calculator has a TABLE feature that enables you to easily construct tables of 
points that satisfy the given equation. 



3. 1 . GRAPHING EQUATIONS BY HAND 



161 



EXAMPLE 4. Use the graphing calculator to help create a table of points 
that satisfy the equation y = x 2 — 7 . Plot the points in your table. If you don't 
feel that there is enough evidence to envision what the final shape of the graph 
will be, use the calculator to add more points to your table and plot them. 
Continue this process until your are convinced of the final shape of the graph. 

Solution: The first step is to load the equation y = x 2 — 7 into the Y= menu 
of the graphing calculator. The topmost row of buttons on your calculator (see 
Figure 3.10) have the following appearance: 



You Try It! 



TRACE 




GRAPH 



WINDOW 



Pressing the Y= button opens the Y= menu shown in Figure 3.11. Use the 
up-and-down arrow keys (see Figure 3.13) to move the cursor after Yl= in the 
Y= menu, then use the following keystrokes to enter the equation y = x 2 — 7. 
The result is shown in Figure 3.12. 



mom 



ENTER 



Ploti 


Plots 


Plot3 


\Vl = 






■^E = 






■^3 = 






\Yh = 






■^Ye = 






^Vfi = 






^v?= 







Ploti Plots Plots 


^iBX A 2-7 


■^£=1 




■^3 = 




\Yh = 




■^Ye = 




^Vfi = 




sV? = 





Figure 3.11: Opening the Y= 
menu. 



Figure 3.12: Entering the equation 

y = x 2 - 7. 




Figure 3.10: The graphing 
calculator. 



The next step is to "set up" the table. First, note that the calculator has 
symbolism printed on its case above each of its buttons. Above the WINDOW 
button you'll note the phrase TBLSET. Note that it is in the same color as the 
2ND button. Thus, to open the setup window for the table, enter the following 
keystrokes. 



WINDOW 



Set TblStart (see Figure 3.14) equal to the first x- value you want to see in 
your table. In this case, enter —4 after TblStart. Set ATbl to the increment 
you want for your x- values. In this case, set ATbl equal to 1. Finally, set 




Figure 3.13: Arrow keys move 
the cursor. 



162 



CHAPTER 3. INTRODUCTION TO GRAPHING 



both the independent and dependent variables to "automatic." In each case, 
use the arrow keys to highlight the word AUTO and press ENTER. The result 
is shown in Figure 3.14. 

Next, note the word TABLE above the GRAPH button is in the same color 
as the 2ND key. To open the TABLE, enter the following keystrokes. 



GRAPH 



The result is shown in Figure 3.15. Note that you can use the up-and-down 
arrow keys to scroll through the table. 



TABLE SETUP 

TblStart=-4 

*Tbl=I 

Indpnt: EH135 flsk 
Depend: rslffiK flsk 



Figure 3.14: Opening the TBLSET 
menu. 



X 


Vi 




-i 



1 


9 
£ 

-fi 
-? 
-fi 
"3 




Press + for ^Tbll 



Figure 3.15: Table of points satis- 
fying y = x 2 - 7. 



Next, enter the results from your calculator's table into a table on a sheet 
of graph paper, then plot the points in the table. The results are shown in 
Figure 3.16. 

V 



X 


y = x 2 - 7 


(x,y) 


-4 


9 


(-4,9) 


-3 


2 


(-3,2) 


-2 


-3 


(-2,-3) 


-1 


-6 


(-1,-6) 





-7 


(0,-7) 


1 


-6 


(1,-6) 


2 


-3 


(2,-3) 


3 


2 


(3,2) 


4 


9 


(4,9) 



















11 


) 


. 
















































































































































































































































































































































































































1( 






































1 































































































































































































































































































































































1( 


) 


1 























Figure 3.16: Plotting nine points that satisfy the equation y = x 2 — 7. 



In Figure 3.16, the eventual shape of the graph of y 



7 may be evident 



already, but let's add a few more points to our table and plot them. Open the 



3. 1 . GRAPHING EQUATIONS BY HAND 



163 



table "setup" window again by pressing 2ND WINDOW. Set TblStart to -4 
again, then set the increment ATbl to 0.5. The result is shown in Figure 3.17. 



TABLE SETUP 

TblStart=-4 

*Tbl=.5 

Indpnt: [ilBE flsk 
Depend: 



Figure 3.17: Set ATbl equal to 0.5. 



X 


V1 




-S.E 
-3 

-£.£ 
-E 
-i.E 
-i 


9 

£.££ 

2 

-.?s 

"3 

"H.FE 

-fi 






ku=h 



Figure 3.18: More points satisfying 
y = x 2 - 7. 



Add these new points to the table on your graph paper and plot them (see 
Figure 3.19). 



X 


y = x 2 - 7 


(x,y) 


-3.5 


5.25 


(-3.5,5.25) 


-2.5 


-0.75 


(-2.5,-0.75) 


-1.5 


-4.75 


(-1.5,-4.75) 


-0.5 


-6.75 


(-0.5,-6.75) 


0.5 


-6.75 


(0.5,-6.75) 


1.5 


-4.75 


(1.5,-4.75) 


2.5 


-0.75 


(2.5,-0.75) 


3.5 


5.25 


(3.5,5.25) 



















1( 


M 


































































































































































• 














• 


































































































































































































































If 














• 










• 














10 










































































































































• 






• 
















































































if 


p 


























































































































1f 


)■ 

























Figure 3.19: Adding additional points to the graph of y 



There are an infinite number of points that satisfy the equation y = x 2 — 7. In 
Figure 3.19, we've plotted only 17 points that satisfy the equation y = x 2 — 7. 
However, the collection of points in Figure 3.19 suggest that if we were to plot 
the remainder of the points that satisfy the equation y = x 2 — 7, the result 
would be the curve (called a parabola) shown in Figure 3.20. 



164 



CHAPTER 3. INTRODUCTION TO GRAPHING 



10 



y = x 2 - 7 



-► x 
10 



-10- 



m 



-10- 



Figure 3.20: The graph of y = x 2 — 7 is a curve called a parabola. 



□ 



3.1. GRAPHING EQUATIONS BY HAND 



165 



**>**■**> Exercises ■** * «•* 



In Exercises 1-6, set up a Cartesian Coordinate system on a sheet of graph paper, then plot the given 
points. 



1. A(2,-4), B(-4,-3), and C(-3,2) 

2. 4(3,-4), 5(-3,-2), and C(-4,4) 

3. 4(3,-2), 5(-3,-4), and C(-2, 3) 



4. 4(3,-4), B(-3,-4), and C(-4,2) 

5. A(-2,2), 5(-2,-3), and C(2, -2) 

6. A(-3,2), 5(2,-4), and C(-3, -4) 



In Exercises 7-10, identify the coordinates of point P in each of the given coordinate systems. 
7. 9. 



y 



10. 



166 



CHAPTER 3. INTRODUCTION TO GRAPHING 



11. Which of the points (8, 36), 
and (—2, —12) is a solution 
y = 5x — 5? 

12. Which of the points (0, 
(1,-14), and (-6,36) is a 
equation y = — 7x — 7? 

13. Which of the points (- 
(-6,39), and (2,-3) is a 
equation y = — 5x + 6? 

14. Which of the points (7, - 
(-6,19), and (5,-11) is a 
equation y = — 3a; + 3? 



(5,20), (4,13), 
of the equation 

-9), (6,-46), 
solution of the 



-9,49), (1,1), 
solution of the 



-15), (-9,30), 
solution of the 



15. Which of the points (1,12), (7,395), 
(—7,398), and (0,1) is a solution of the 
equation y = 8x 2 + 3? 

16. Which of the points (-7,154), (7,153), 
(—2,21), and (2,16) is a solution of the 
equation y = 3a; 2 + 6? 

17. Which of the points (8,400), (5,158), 
(0, 3), and (2, 29) is a solution of the equa- 
tion y = 6a; 2 + 2ai? 

18. Which of the points (-6, -114), (6, -29), 
(—7,-149), and (4,-1) is a solution of 
the equation y = —2a; 2 + 7a;? 



In Exercises 19-26, complete the table of points that satisfy the given equation. Set up a coordinate 
system on a sheet of graph paper, plot the points from the completed table, then draw the graph of 
the given equation. 



19. 



X 


y = 2x + 1 


(x,y) 


-3 

-2 
-1 



1 

2 

3 







21. 



X 


y=\x-5\ 


(x,y) 


2 
3 

4 
5 
6 

7 
8 







20. 



X 


y = x — 4 


(x,y) 


-3 

-2 
-1 



1 

2 

3 







22. 



X 


y=\x + 2\ 


(x,y) 


-5 
-4 
-3 
-2 
-1 



1 







3.1. GRAPHING EQUATIONS BY HAND 



167 



23. 



24. 



X 


y = {x + l) 2 


{x,y) 


-4 
-3 
-2 
-1 



1 

2 







25. 



X 


2/=(z + 5) 2 


(z,y) 


-8 
-7 
-6 
-5 
-4 
-3 
-2 







26. 



X 


y = —x — 5 


(a, 3/) 


-3 
-2 
-1 



1 

2 

3 







a; 


j/= -2a; + 3 


(x,y) 


-3 
-2 
-1 



1 

2 

3 







27. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = x 2 — 6x + 5. Use integer values of 
x, starting at and ending at 6. After 
completing the table, set up a coordinate 
system on a sheet of graph paper. Label 
and scale each axis, then plot the points 
in your table. Finally, use the plotted 
points as evidence to draw the graph of 



29. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = — x 2 + 2x + 3. Use integer values of 
X, starting at —2 and ending at 4. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 



y 



6x + 5. 



of y 



2x + 3. 



28. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = x 2 — 2x — 3. Use integer values of 
x, starting at —2 and ending at 4. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 



30. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = —x 2 — 2x + 8. Use integer values of 
x, starting at —5 and ending at 3. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 



of y 



2a; -3. 



of y 



2x 



168 



CHAPTER 3. INTRODUCTION TO GRAPHING 



31. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = x 3 — 4a; 2 — 7x + 10. Use integer val- 
ues of x, starting at —3 and ending at 6. 
After completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 



of y 



32. 



4a; 2 



7x + 10. 



Use a graphing calculator to complete 
a table of points satisfying the equation 
y = x 3 + 3x 2 — 13a; — 15. Use integer val- 
ues of x, starting at —6 and ending at 4. 
After completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 



of y 



3a; 2 - 13a; - 15. 



33. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = —x 3 + Ax 2 + 7x — 10. Use integer val- 
ues of x, starting at —3 and ending at 6. 
After completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 
of y = -x 3 + Ax 2 + 7x- 10. 



34. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = —x 3 + x 2 + 12x. Use integer values 
of x, starting at —4 and ending at 5. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 
of y = -x 3 + x 2 + Ylx. 

35. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = \/x + 5. Use integer values of x, 
starting at —5 and ending at 10. Round 
your y-values to the nearest tenth. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 
of y = \/x + 5. 

36. Use a graphing calculator to complete 
a table of points satisfying the equation 
y = \/4 — x. Use integer values of x, 
starting at —10 and ending at 4. Round 
your y-values to the nearest tenth. Af- 
ter completing the table, set up a coor- 
dinate system on a sheet of graph paper. 
Label and scale each axis, then plot the 
points in your table. Finally, use the plot- 
ted points as evidence to draw the graph 
of y = i/4 — x. 



j*- j*- j*- Answers •*$ •** ■*$ 



1. 













5 






































c(- 


3,2) 






































































5 


















r 




















































B(-4, - 


3) L 




























5 






A 


% 


-] 


) 



C(-2,3) 



£(-3,-4). 



A(3, -2) 



3. 1 . GRAPHING EQUATIONS BY HAND 



169 



23. 













5^ 


































J 


l(-2,L'l 






























































5 


















5 








































, C*(2,'-2) 


B{ 


-2, -3)" 
























5- 













7. (-4,3) 
9. (3,-4) 
11. (5,20) 
13. (1,1) 
15. (7,395) 
17. (8,400) 
19. 



-10 



10 



y y = 2x + 1 



-10 



10 



21. 



















L( 






















































































































































J/ = 






























































































































































































































































, 










































11 






































i 



























































































































































































































































































































r 



























y = \x- 5| 



25. 



27. 



y=(x + l) 2 



-10 



-10 



104 



§ 



-io4 



10 



















L( 


I 


/ 






















































































































































































































































































































































































































































r 










































* 


1( 


) 




































lu 
























































































































































































































































































































( 


' 

























y = —x — 5 



§ 



-y = x — dx + 5 



10 



170 



CHAPTER 3. INTRODUCTION TO GRAPHING 



29. 



-io- 



33. 



10 



I 



10 



-10 



y = -x 2 + 2x + 3 



y 

50 + 



-50 



r- 






10 



-y = -x s + Ax 2 + 7x - 10 



ltd 



31. 



-io- 



35. 



-50 



A 



-50 



^! 



-y = X 3 - ix 2 - 7x + 10 



10 



















LC 


V 






























































































































































































































— , 




























































































































































































































































' 


10 




































lu 
























































































































































































































































































































( 


<■ 























3.2. THE GRAPHING CALCULATOR 



171 



3.2 The Graphing Calculator 

It's time to learn how to use a graphing calculator to sketch the graph of an 
equation. We will use the TI-84 graphing calculator in this section, but the 
skills we introduce will work equally well on the ancient TI-82 and the less 
ancient TI-83 graphing calculators. 

In this introduction to the graphing calculator, you will need to use several 
keys on the upper half of the graphing calculator (see Figure 3.21). The up- 
and-down and left-and-right arrow keys are located in the upper right corner 
of Figure 3.21. These are used for moving the cursor on the calculator view 
screen and various menus. Immediately below these arrow keys is the CLEAR 
button, used to clear the view screen and equations in the Y= menu. 



s-wpldi n »»„ n r0 n*u,j n caic f* 



o o o yM 

'-^ "N" HIT ^— 

^^^ ^^^ ^^^ 




Figure 3.21: Top half of calculator. 



It is not uncommon that people share their graphing calculators with 
friends who make changes to the settings in the calculator. Let's take a moment 
to make sure we have some common settings on our calculators. 

Locate and push the MODE in the first row of Figure 3.21. This will open 
the window shown in Figure 3.22. Make sure that the mode settings on your 
calculator are identical to the ones shown in Figure 3.22. If not, use the up-and- 
down arrow keys to move to the non-matching line item. This should place a 
blinking cursor over the first item on the line. Press the ENTER button on the 
lower- right corner of your calculator to make the selection permanent. Once 
you've completed your changes, press 2ND MODE again to quit the MODE 
menu. 

Next, note the buttons across the first row of the calculator, located imme- 
diately below the view screen. 



TRACE 




GRAPH 



Above each button are one 
or more commands located 
on the calculator's case. 
Press the 2ND key to access 
a command having the same 
color as the 2ND key. Press 
the ALPHA key to access a 
command having the same 
color as the ALPHA key. 



The QUIT command is 
located above the MODE 
button on the calculator case 
and is used to exit the 
current menu. 



172 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Just above the Y= button are the words STAT PLOT, printed in the same color 
as the 2ND button located near the top left of the calculator (see Figure 3.21). 
Press the 2ND button, then the Y= button. This opens the stat plot menu 
shown in Figure 3.23. 



FLOAT 
RhDIhTi 

nine 

cannECTED 

SEQUERTIAL 
REAL 



sci Eni] 

0i234£fi7B9 
DEGREE 
PAR PDL SEQ 
DDT 
SIHUL 

HORIZ G-T 
+TIEKT -I- 



Figure 3.22: Settings in the MODE 
window. 



afflfldcqp 


yHF , To€i..j!!rff 


L-1L1 


L2 d 


2:Plot2. 


..Off 


LHL1 


L2 a 


3:Plot3. 


..Off 


LHL1 


L2 d 


4-1-PlotsOff 



Figure 3.23: The STAT PLOT 
menu. 



You Try It! 



Use the graphing calculator 
to sketch the graph of 
y = —x + 3. 



We need all of the stat plots to be "off." If any of the three stat plots are 
"on," select 4:PlotsOff (press the number 4 on your keyboard), then press 
the ENTER key on the lower right corner of your calculator. 

That's it! Your calculator should now be ready for the upcoming exercises. 



EXAMPLE 1. Use the graphing calculator to sketch the graph of y 



1. 



Solution: Recall that we drew the graph of y = x + 1 by hand in Example 1 
of Section 3.1 (see Figure 3.9). In this example, we will use the graphing cal- 
culator to produce the same result. 

Press the Y= button. The window shown in Figure 3.24 appears. If any 
equations appear in the Y= menu of Figure 3.24, use the up-and-down arrow 
keys and the CLEAR button (located below the up-and-down and left-and- 
right arrow keys) to delete them. 



Ploti 


Plots 


Plot3 


Wi = 






Wz = 






■^3 = 






\Yh = 






■^Ye = 






^Vg = 






"^7 = 







Figure 3.24: Press the Y= button 
to open the Y= menu. 



Ploti 


Plots 


Plots 


WlBX+1 




sV2 = 






\Vl = 






\Yh = 






■^Ye = 






^Vg = 






^7 = 







Figure 3.25: Enter y 
Yl=. 



1 in 



Next, move the cursor to Yl = , then enter the equation y = x + 1 in Yl with 
the following button keystrokes. The result is shown in Figure 3.25. 



3.2. THE GRAPHING CALCULATOR 



173 



□ 



ENTER 



Next, select the ZOOM key on the top row of the calculator. This will open the 
window shown in Figure 3.26. From the ZOOM menu, select 6:ZStandard. 
There are two ways to make this selection: 

1. Use the down-arrow key to move downward in the ZOOM menu until 
6:ZStandard is highlighted, then press the ENTER key. 

2. A quicker alternative is to simply press the number 6 on the calculator 
keyboard to select 6:ZStandard. 

The resulting graph of y = x + 1 is shown in Figure 3.27. Note that the result is 
identical to the graph of y = x + 1 drawn by hand in Example 3 of Section 3.1 
(see Figure 3.9). 



MEMORY 
ox 
:Zoon In 
3:Zoon Out 
4:ZDecinal 
5:ZS=iuare 
6:ZStandard 
7-i-ZTrig 




Figure 3.26: Select 6:ZStandard 
from the ZOOM menu. 



Figure 3.27: The graph of y = x- 
is a line. 



Answer: 




□ 



EXAMPLE 2. Use the graphing calculator to sketch the graph of y 



-7. 



Solution: Recall that we drew the graph of y = x 2 — 7 by hand in Example 4 
of Section 3.1 (see Figure 3.20). In this example, we will use the graphing 
calculator to produce the same result. 



Press the Y= button and enter the equation y 
Figure 3.28) with the following keystrokes: 



7 into Yl (see 



X,T,9,n 



mam 



ENTER 



You Try It! 



Use the graphing calculator 
to sketch the graph of 
y = -x 2 + 4. 



The caret (A) symbol (see Figure 3.21) is located in the last column of buttons 
on the calculator, just underneath the CLEAR button, and means "raised to." 
The caret button is used for entering exponents. For example, x 2 is entered as 



XA2, 



is entered as XA3, and so on. 



174 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Answer: 



Press the ZOOM button, then select 6:ZStandard from the ZOOM menu 
to produce the graph of y = x 2 — 7 shown in Figure 3.29. Note that the result in 
Figure 3.29 agrees with the graph of y = x 2 — 7 we drew by hand in Example 4 
of Section 3.1 (see Figure 3.20). 




Ploti Plots Plots 


^13X^2-7 


■^£=1 




\Vi = 




\Yh = 




■^Ye = 




^Vfi = 




\Y? = 






Figure 3.28: Enter y 
Yl. 



7 in 



Figure 3.29: The graph of y 
is a parabola. 



□ 



Reproducing Calculator Results on Homework Paper 

In this section we delineate recommendations and requirements when repro- 
ducing graphing calculator results on your homework paper. 

Consider again the final result of Example 2 shown in Figure 3.29. To 
determine the scale at each end of each axis, press the WINDOW button on 
the topmost row of buttons on your calculator. The WINDOW settings for 
Figure 3.29 are shown in Figure 3.30. Figure 3.31 presents a visual explanation 
of each of the entries Xmin, Xmax, Ymin, and Ymax. 



WINDOW 
Xnun=-10 
Xnax=l@ 
Xscl=l 
Wiin=-10 
Wiax=l@ 
Vscl=l 

■i-Xres=l 



Figure 3.30: The WINDOW pa- 
rameters. 



y 

10 f Ymax 



Xmin 



-10 



-10 



Xmax 

x 



10 



Ymin 



Figure 3.31: Xmin, Xmax, 
Ymin, and Ymax contain the 
scale at the end of each axis. 



3.2. THE GRAPHING CALCULATOR 



175 



Xmin and Xmax indicate the scale at the left- and right-hand ends of the 
a;-axis, respectively, while Ymin and Ymax indicate the scale at the bottom- 
and top-ends of the y-axis. Finally, as we shall see shortly, Xscl and Yscl 
control the spacing between tick marks on the x- and j/-axes, respectively. 

When reproducing the graph in your calculator viewing window on your 
homework paper, follow the Calculator Submission Guidelines. 

Calculator Submission Guidelines. 

1. All lines (including the horizontal and vertical axes) should be drawn 
with a ruler. All curves should be drawn freehand. 

2. Set up a coordinate system on your homework paper that mimics closely 
the coordinate system in your calculator's view screen. Label your axes 
(usually with x and y). 

3. Indicate the scale at each end of each axis. Use Xmin, Xmax, Ymin, 
and Ymax in the WINDOW menu for this purpose. 

4. Copy the graph from your calculator's viewing window onto your coordi- 
nate system. Label the graph with its equation. 



For example, to report the results of Figure 3.29, draw the axes with a ruler, 
label the horizontal axis with x, the vertical axis with y, then place the values 
of Xmin, Xmax, Ymin, and Ymax at the appropriate end of each axis (see 
Figure 3.32). 




Figure 3.32: Reporting the graph of y = x 2 — 7 on your homework paper. 



Because the graph is a curve, make a freehand copy that closely mimics 
the graph shown in Figure 3.29, then label it with its equation, as shown in 



176 



CHAPTER 3. INTRODUCTION TO GRAPHING 



You Try It! 



Approximate the point of 
intersection of the graphs of 
y = x — 4 and y = 5 — x 
using the TRACE button. 



Figure 3.32. 

Adjusting the Viewing Window 

Note that Figure 3.31 gives us some sense of the meaning of the "standard 
viewing window" produced by selecting 6:ZStandard from the ZOOM menu. 
Each time you select 6:ZStandard from the ZOOM menu, Xmin is set to —10, 
Xmax to 10, and Xscl to 1 (distance between the tick marks on the x-axis). 
Similarly, Ymin is set to —10, Ymax to 10, and Yscl to 1 (distance between 
the tick marks on the y-axis). You can, however, override these settings as we 
shall see in the next example. 



EXAMPLE 3. Sketch the graphs of the following equations on the same 
viewing screen. 

5 , 2 

y = —x — 3 and y = —x + 4 
y 4 y 3 

Adjust the viewing window so that the point at which the graphs intersect 

(where the graphs cross one another) is visible in the viewing window, then use 

the TRACE button to approximate the coordinates of the point of intersection. 

Solution: We must first decide on the proper syntax to use when entering the 
equations. In the case of the first equation, note that 



y 



is pronounced "y equals five-fourths x minus three," and means "y equals five- 
fourths times x minus three." Enter this equation into Yl in the Y= menu as 
5/4*X— 3 (see Figure 3.33), using the button keystrokes: 



0H0 







The division and times buttons are located in the rightmost column of the 
calculuator. Similarly, enter the second equation into Y2 as 2/3*X+4 using the 
button keystrokes: 



0B0 



A, T, 9, n 



□ 



Select 6:ZStandard in the ZOOM menu to produce the lines in Figure 3.34. 
When we examine the resulting graphs in Figure 3.34, it appears that their 
point of intersection will occur off the screen above and to the right of the 
upper right-corner of the current viewing window. With this thought in mind, 
let's extend the a;-axis to the right by increasing the value of Xmax to 20. 
Further, let's extend the y-axis upward by increasing the value of Ymin to 20. 
Press the WINDOW button on the top row of the calculator, then make the 
adjustments shown in Figure 3.35. 



3.2. THE GRAPHING CALCULATOR 



177 



Ploti Plots Plots 
WiB5^4*tf-3 
^eB2^3*X+4 

^7 = 




Figure 3.33: Enter equations. 



Figure 3.34: The graphs. 



WINDOW 
Xnun=-10 
Xnax=2@ 
Xscl=l 
Vmin=-10 
Wiax=20 
Vscl=l 

■i-Xres=l 



. . . . .^^. . . 



Figure 3.35: Adjusting the viewing 
window. 



Figure 3.36: The point of intersec- 
tion is visible in the new viewing 
window. 



If we select 6:ZStandard from the ZOOM menu, our changes to the WIN- 
DOW parameters will be discarded and the viewing window will be returned to 
the "standard viewing window." If we want to keep our changes to the WIN- 
DOW parameters, the correct approach at this point is to push the GRAPH 
button on the top row of the calculator. The resulting graph is shown in 
Figure 3.36. Note that the point of intersection of the two lines is now visible 
in the viewing window. 

The image in Figure 3.36 is ready for recording onto your homework. How- 
ever, we think we would have a better picture if we made a couple more changes: 

1. It would be nicer if the point of intersection were more centered in the 
viewing window. 

2. There are far too many tick marks. 

With these thoughts in mind, make the changes to the WINDOW parame- 
ters shown in Figure 3.37, then push the GRAPH button to produce the image 
in Figure 3.38. 



178 



CHAPTER 3. INTRODUCTION TO GRAPHING 



WINDOW 
Xnin=l5 
Xnax=25 
Xscl=5 
Ynun= - 5 
Wiax=25 
Vscl=5 

■i-Xres=l 




Figure 3.37: A final adjustment of 
the viewing window. 



Figure 3.38: A centered point of in- 
tersection and fewer tick marks. 



The TRACE button is only 
capable of providing a very 
rough approximation of the 
point of intersection. In 
Chapter 4, we'll introduce 
the intersection utility on the 
CALC menu, which will 
report a much more accurate 
result. 



Answer: (4.68,0.68) 



Y1=K-H"-v 




K=1.fiH0HEli 


V=.fiH0HE10fi 



Finally, push the TRACE button on the top row of buttons, then use the 
left- and right-arrow keys to move the cursor atop the point of intersection. 
An approximation of the coordinates of the point of intersection are reported 
at the bottom of the viewing window (see Figure 3.39). 



V1=E/H*K-3 




K=ii.91HB9H Y=ii.H93fii7 



Figure 3.39: Approximate coordi- 
nates of the point of intersection re- 
ported by the TRACE button. 




Figure 3.40: Reporting the answer 
on your homework. 



In reporting the answer on our homework paper, note that we followed the 
Calculator Submission Guidelines on page 175. 

□ 



3.2. THE GRAPHING CALCULATOR 179 



**. i*. **. Exercises ■*$*$-*$ 

In Exercises 1-16, enter the given equation in the Y= menu of your calcuator, then select 6:ZStandard 
from the ZOOM menu to produce its graph. Follow the Calculator Submission Guidelines outlined in 
the subsection Reproducing Calculator Results on Homework Paper when submitting your result on 
your homework paper. 

1. y = 2x + 3 

2. y = 3- 2x 

3. y = -x - 4 
y 2 

2 

4. y = --x + 2 
y 3 

5. y = 9-x 2 

6. y = x 2 -4 

7. y = -x 2 - 3 
y 2 

8. h = 4 x 2 

y 3 



9. 


y = 


\fx 


10. 


y = 
y = 
y = 


^-x 


11. 


Vx + 3 


12. 


\/5 — x 


13. 


y = 


|*| 


14. 


y = 


-N 


15. 


y = 


|» + 2| 


16. 


y = 


-|a: -3 



In Exercises 17-22, the graph of each given equation is a parabola, having a "u-shape" similar to the 
graph in Figure 3.32. Some of the parabolas "open up" and some "open down." Adjust Ymin and/or 
Ymax in the WINDOW menu so that the "turning point" of the parabola, called its vertex, is visible in 
the viewing screen of your graphing calculator. Follow the Calculator Submission Guidelines outlined 
in the subsection Reproducing Calculator Results on Homework Paper when submitting your result on 
your homework paper. 



17.2/ = 


x 2 + x - 20 


18. y = 


-x 2 - 2a; + 24 


19. y = 


-x 2 +4x + 21 



20. y = x 2 + 6x - 16 

21. y = 2x 2 - 13x-24 

22. y = -3x 2 - 19a; + 14 



180 



CHAPTER 3. INTRODUCTION TO GRAPHING 



In Exercises 23-28, sketch the given lines, then adjust the WINDOW parameters so that the point 
of intersection of the two lines is visible in the viewing window. Use the TRACE key to estimate 
the point of intersection. Follow the Calculator Submission Guidelines outlined in the subsection 
Reproducing Calculator Results on Homework Paper when submitting your result on your homework 
paper. Be sure to also include your estimate of the point of intersection in your sketch. 



23. y = 2x + 1 and y = x + ; 

24. y = 4 — 2x and y = — 6 - 

1 2 

25. y = —x and y = —x — 3 
y 2 J 3 



26. y = 2x + 5 and y 



= x — 



27. y 

28. y 



= 5 — x and y 
1 

2 



= -3- 2x 

1 , 4 
= 1 x and y = —4 x 

2 5 



J*. t»- ;*■ 



Answers 



•*;■*; •*! 



y = 2t + 3 






10 


. 


-10 


' —10 


\ 10 

1 



2/ = 9 - x 2 



3. 



y=2 x 




12 o 





. 10' 


. 


-10 


-10 


/ 10 

■ 



3.2. THE GRAPHING CALCULATOR 



181 



10 



-10 



-10 



y = v^ 



10 



15. 



y y-\x + 2\ 





10 


- 


-10 


-10 


10 



11. 



17. 



-10 



y — \Jx + 3 




2/ y = x 2 + x - 20 





30 


. 


-10 


-30 


/ 10 

1 



13. 



19. 



V = 1*1 





j/ = -x 2 +4x + 21 



182 



CHAPTER 3. INTRODUCTION TO GRAPHING 



21. 



25. 



y y = 2x 2 - 13a; - 24 





. 50 


, 


-10 


-50 


/ 10 



y = 9 x 




(18.08,9.04) 



35 



23. 



y = x + 




27. 




y = 2x + l 



y — 5 — x 



y = -3-2x 



3.3. RATES AND SLOPE 



183 



3.3 Rates and Slope 

Let's open this section with an application of the concept of rate. 



EXAMPLE 1. An object is dropped from rest, then begins to pick up speed 
at a constant rate of 10 meters per second every second (10 (m/s)/s or 10 m/s 2 ). 
Sketch the graph of the speed of the object versus time. 

Solution: In this example, the speed of the object depends on the time. This 
makes the speed the dependent variable and time the independent variable. 

Independent versus dependent. It is traditional to place the independent 
variable on the horizontal axis and the dependent variable on the vertical axis. 



You Try It! 



Starting from rest, an 
automobile picks up speed at 
a constant rate of 5 miles per 
hour every second 
(5 (mi/hr)/s). Sketch the 
graph of the speed of the 
object versus time. 



Following this guideline, we place the time on the horizontal axis and the speed 
on the vertical axis. In Figure 3.41, note that we've labeled each axis with the 
dependent and independent variables (v and t), and we've included the units 
(m/s and s) in our labels. 

Next, we need to scale each axis. In determining a scale for each axis, keep 
two thoughts in mind: 

1. Pick a scale that makes it convenient to plot the given data. 

2. Pick a scale that allows all of the given data to fit on the graph. 

In this example, we want a scale that makes it convenient to show that the 
speed is increasing at a rate of 10 meters per second (10 m/s) every second (s). 
One possible approach is to make each tick mark on the horizontal axis equal 
to 1 s and each tick mark on the vertical axis equal to 10 m/s. 



v(m/s) 

100 

90 
80 
70 
CO 
50 
40 
30 
20 
10 



v(m/s) 

100 

90 
80 
70 
(30 
50 
40 
30 
20 
10 



i(s) 



0123456789 10 

Figure 3.41: Label and scale each 
axis. Include units with labels. 







' I I I I I I I I I t 
<> — 

— ii — 

ii 

ii 

ii 

ii 

ii 

ii 

— ii 

n — I — I — I — I — I — I — I — I — I — L> 



0123456789 10 



t(s) 



Figure 3.42: Start at (0,0), then 
continuously move 1 right and 10 
up. 



184 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Next, at time t = s, the speed is v = m/s. This is the point (t, v) = (0, 0) 
plotted in Figure 3.42. Secondly, the rate at which the speed is increasing is 10 
m/s per second. This means that every time you move 1 second to the right, 
the speed increases by 10 m/s. 

In Figure 3.42, start at (0,0), then move 1 s to the right and 10 m/s up. 
This places you at the point (1, 10), which says that after 1 second, the speed 
of the particle is 10 m/s. Continue in this manner, continuously moving 1 s to 
the right and 10 m/s upward. This produces the sequence of points shown in 
Figure 3.42. Note that this constant rate of 10 (m/s)/s forces the graph of the 
speed versus time to be a line, as depicted in Figure 3.43. 



Answer: 



u(mi/hr) 




i>(m/s) 
100 




0123456789 10 
Figure 3.43: The constant rate forces the graph to be a line. 



□ 



Measuring the Change in a Variable 

To calculate the change in some quantity, we take a difference. For example, 
suppose that the temperature in the morning is 40° F, then in the afternoon 
the temperature measures 60° F (F stands for Fahrenheit temperature). Then 
the change in temperature is found by taking a difference. 

Change in temperature = Afternoon temperature — Morning temperature 
= 60° F - 40° F 
= 20° F 



Therefore, there was a twenty degree increase in temperature from morning to 
afternoon. 



3.3. RATES AND SLOPE 185 

Now, suppose that the evening temperature measures 50° F. To calculate 
the change in temperature from the afternoon to the evening, we again subtract. 

Change in temperature = Evening temperature — Afternoon temperature 
= 50°F-60°F 
= -10° F 

There was a ten degree decrease in temperature from afternoon to evening. 

Calculating the Change in a Quantity. To calculate the change in a 
quantity, subtract the earlier measurement from the later measurement. 

Let T represent the temperature. Mathematicians like to use the symbolism 
AT to represent the change in temperature. For the change in temperature 
from morning to afternoon, we would write AT = 20° F. For the afternoon to 
evening change, we would write AT = —10° F. 

Mathematicians and scientists make frequent use of the Greek alphabet, 
the first few letters of which are: 

a, j3, 7, d, . . . (Greek alphabet, lower case) 

A, B,T,A,... (Greek alphabet, upper case) 

a, b,c,d, . . . (English alphabet) 

Thus, the Greek letter A, the upper case form of 5, correlates with the letter 'rf' 
in the English alphabet. Why did mathematicians make this choice of letter to 
represent the change in a quantity? Because to find the change in a quantity, 
we take a difference, and the word "difference" starts with the letter 'd.' Thus, 
AT is also pronounced "the difference in T." 

Important Pronunciations. Two ways to pronounce the symbolism AT. 

1. AT is pronounced "the change in T." 

2. AT is also pronounced "the difference in T." 



Slope as Rate 

Here is the definition of the slope of a line. 

Slope. The slope of a line is the rate at which the dependent variable is chang- 
ing with respect to the independent variable. For example, if the dependent 



186 



CHAPTER 3. INTRODUCTION TO GRAPHING 



variable is y and the independent variable is x, then the slope of the line is: 

Ay 



Slope 



Ax 



You Try It! 



Starting from rest, an 
automobile picks up speed at 
a constant rate of 5 miles per 
hour every second 
(5 (mi/hr)/s). The constant 
rate forces the graph of the 
speed of the object versus 
time to be a line. Calculate 
the slope of this line. 



EXAMPLE 2. In Example 1, an object released from rest saw that its speed 
increased at a constant rate of 10 meters per second per second (10 (m/s)/s or 
10m/s 2 ). This constant rate forced the graph of the speed versus time to be a 
line, shown in Figure 3.43. Calculate the slope of this line. 

Solution: Start by selecting two points -P(2,20) and Q(8,80) on the line, as 
shown in Figure 3.44. To find the slope of this line, the definition requires that 
we find the rate at which the dependent variable v changes with respect to the 
independent variable t. That is, the slope is the change in v divided by the 
change in t. In symbols: 

Slope = — — 
1 At 



w(m/s) 

100 

90 
80 
70 
60 
50 
40 
30 
20 
10 




. 




























1-- 


)(8 


■ a 


- 1 J 






























































































































p 


(•>■ 


9f 


) 



























12 3 4 5 6 7 



9 10 



t( S ) 



u(m/s) 

100 
90 

80 
70 
60 

50 
40 
30 
20 
10 




. 














































Q 


)(7, to) 






























































































p 


(3 3r 


) 

















































0123456789 10 



too 



Figure 3.45: The slope does not de- 
pend on the points we select on the 
line. 



Figure 3.44: Pick two points to 
compute the slope. 



Now, as we move from point P(2, 20) to point Q(8, 80), the speed changes from 
20 m/s to 80 m/s. Thus, the change in the speed is: 



Aw = 80 m/s- 20 m/s 
= 60 m/s 



3.3. RATES AND SLOPE 187 



Similarly, as we move from point P(2,20) to point Q(8, 80), the time changes 
from 2 seconds to 8 seconds. Thus, the change in time is: 

A^ = 8s- 2s 
= 6s 

Now that we have both the change in the dependent and independent variables, 
we can calculate the slope. 

Slope = — — 
At 

60m/s 
~ 6s 

= 10^ 

s 

Therefore, the slope of the line is 10 meters per second per second (10 (m/s)/s 
or 10m/s 2 ). 

The slope of a line does not depend upon the points you select. Let's try 
the slope calculation again, using two different points and a more compact 
presentation of the required calculations. Pick points P(3,30) and Q(7, 70) as 
shown in Figure 3.45. Using these two new points, the slope is the rate at which 
the dependent variable v changes with respect to the independent variable t. 



Ql AV 

Slope = — — 
At 




70 m/s - 


- 30 m/s 


7s- 


-3s 


40 m/s 




4s 




m/s 
= 10 — 

s 





Again, the slope of the line is 10 (m/s)/s. Answer: 5 (mi/hr)/s 
□ 

Example 2 points out the following fact. 

Slope is independent of the selected points. It does not matter which 
two points you pick on the line to calculate its slope. 

The next example demonstrates that the slope is also independent of the 
order of subtraction. 



188 



CHAPTER 3. INTRODUCTION TO GRAPHING 



You Try It! 



Compute the slope of the 
line passing through the 
points P(-3, 1) and <5(2, 4). 



EXAMPLE 3. Compute the slope of the line passing through the points 
P(-l,-2) and (3(3,3). 

Solution: First, sketch the line passing through the points P(— 1, — 2) and 
Q(3, 3) (see Figure 3.46). 



Warning! If you are not 

consistent in the direction 
you subtract, you will not 
get the correct answer for 
the slope. For example: 



(-2) 



-1 



In this case, we subtracted 
the y-coordinate of point 
P(-l,-2) from the 
y-coordinate of point Q(3, 3), 
but then we changed horses 
in midstream, subtracting 
the x-coordinate of point 
(5(3, 3) from the 
cc-coordinate of point 
P(-l, -2). Note that we get 
the negative of the correct 
answer. 




Figure 3.46: Computing the slope of the line passing through the points 
P(-l,-2) and Q(3,3). 

To calculate the slope of the line through the points P(— 1, —2) and <5(3, 3), we 
must calculate the change in both the independent and dependent variables. 
We'll do this in two different ways. 



Subtract the coordinates of point 
P(— 1,— 2) from the coordinates of 
point (5(3, 3). 



Slope = 


_ Ay 
Air 






3-(- 


-2) 




3-(- 


-1) 




5 

" 4 





Subtract the coordinates of point 
Q(3,3) from the coordinates of 
point P(-l,-2). 

Slope = — — 

Air 



-1-3 
-5 



Answer: 3/5 



_ 5 
~ 4 
Note that regardless of the direction of subtraction, the slope is 5/4. 



□ 



3.3. RATES AND SLOPE 



189 



Example 3 demonstrates the following fact. 

The direction of subtraction does not matter. When calculating the 
slope of a line through two points P and Q, it does not matter which way you 
subtract, provided you remain consistent in your choice of direction. 



The Steepness of a Line 

We need to examine whether our definition of slope matches certain expecta- 
tions. 

Slope and steepness of a line. The slope of a line is a number that tells us 
how steep the line is. 



If slope is a number that measures the steepness of a line, then one would 
expect that a steeper line would have a larger slope. 



EXAMPLE 4. Graph two lines, the first passing through the points P(— 3, — 2) 
and Q(3,2) and the second through the points R(— 1,— 3) and 5(1,3). Calcu- 
late the slope of each line and compare the results. 

Solution: The graphs of the two lines through the given points are shown, 
the first in Figure 3.47 and the second in Figure 3.48. Note that the line in 
Figure 3.47 is less steep than the line in Figure 3.48. 

V 

























































S 


(1. 


3) 






































-5 


















5 


L 


-1. 


















R{- 


*) 










































5- 













Figure 3.47: This line is less steep 
than the line on the right. 



Figure 3.48: This line is 
than the line on the left. 



steeper 



Remember, the slope of the line is the rate at which the dependent variable 
is changing with respect to the independent variable. In both Figure 3.47 and 
Figure 3.48, the dependent variable is y and the independent variable is x. 



You Try It! 



Compute the slope of the 
line passing through the 
points P(— 2, —3) and 
Q(2,5). Then compute the 
slope of the line passing 
through the points 
7?(-2,-l) and 5(5,3), and 
compare the two slopes. 
Which line is steeper? 



190 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Answer: The first line has 
slope 2, and the second line 
has slope 4/7. The first line 
is steeper. 



You Try It! 



Compute the slope of the 
line passing through the 
points P(— 3, 3) and 
Q(3,— 5). Then compute the 
slope of the line passing 
through the points R(— 4, 1) 
and 5(4, —3), and compare 
the two slopes. Which line is 
steeper? 



Subtract the coordinates of point 
P(— 3,— 2) from the coordinates of 
point Q(3,2). 



Subtract the coordinates of the 
point R(—l,—3) from the point 

5(1,3). 



Slope of first line = — — 



Ay 

Ax 

2 -(-2) 



Ay 
Slope of second line = — — 

Ax 



(-3) 



(-3) 



1 

G 
2 
3 



(-1) 



Note that both lines go uphill and both have positive slopes. Also, note that 
the slope of the second line is greater than the slope of the first line. This is 
consistent with the fact that the second line is steeper than the first. 

□ 

In Example 4, both lines slanted uphill and both had positive slopes, the 
steeper of the two lines having the larger slope. Let's now look at two lines 
that slant downhill. 



EXAMPLE 5. Graph two lines, the first passing through the points P(— 3, 1) 
and (5(3,-1) and the second through the points R(— 2,4) and 5(2,-4). Cal- 
culate the slope of each line and compare the results. 

Solution: The graphs of the two lines through the given points are shown, 
the first in Figure 3.49 and the second in Figure 3.50. Note that the line in 
Figure 3.49 goes downhill less quickly than the line in Figure 3.50. 
Remember, the slope of the line is the rate at which the dependent variable 
is changing with respect to the independent variable. In both Figure 3.49 and 
Figure 3.50, the dependent variable is y and the independent variable is x. 



Subtract the coordinates of point 
P(— 3, 1) from the coordinates of 
point Q(3, —1). 



Slope of first line = 


Ax 






-1- 


1 




3-(- 


-3) 




-2 






6 






1 






3 





Subtract the coordinates of point 
R(— 2,4) from the coordinates of 
point 5(2,-4). 



Ay 
Slope of second line = — — 

Ax 



-4-4 

r F2) 



3.3. RATES AND SLOPE 



191 











5^ 


i 


































P 


(" 


3. 












i) 






























5 


















5 










Q 


(3 




1) 












































5 













Figure 3.49: This line goes down- 
hill more slowly than the line on the 
right. 











■J 


/ 






























R(- 


2. 


1) 








































































-5 


















5 




































(2. 


-4) 
















S 










■ r >; 


■ 











Figure 3.50: This line goes downhill 
more quickly than the line on the 
left. 



Note that both lines go downhill and both have negative slopes. Also, note 
that the magnitude (absolute value) of the slope of the second line is greater 
than the magnitude of the slope of the first line. This is consistent with the 
fact that the second line moves downhill more quickly than the first. 



Answer: The first line has 
slope —4/3, and the second 
line has slope —1/2. The 
first line is steeper. 



□ 



What about the slopes of vertical and horizontal lines? 



EXAMPLE 6. Calculate the slopes of the vertical and horizontal lines 
passing through the point (2,3). 

Solution: First draw a sketch of the vertical and horizontal lines passing 
through the point (2,3). Next, select a second point on each line as shown in 
Figures 3.51 and 3.52. 



You Try It! 



Calculate the slopes of the 
vertical and horizontal lines 
passing through the point 

(-4,1). 



5t 



■P(-2,3)— Q(2,3) 



Figure 3.51: A horizontal line 
through (2,3). 



'/ 



<-S(2,3) 



*■ x 



fl(2,-3) 



Figure 3.52: A vertical line through 

(2,3). 



192 



CHAPTER 3. INTRODUCTION TO GRAPHING 



The slopes of the horizontal and vertical lines are calculated as follows. 



Subtract the coordinates of the 
point P(— 2,3) from the coordi- 
nates of the point Q(2, 3). 



Slope of horizontal line 



Ay 

Ax 
_3-3 

2~ 


4 




(-2) 



Subtract the coordinates of the 
point (2,-3) from the coordinates 
of the point 5(2,3). 



Slope of vertical line 



Ay 
Ace 
3 ~(-3) 

2-2 
6 

undefined 



Answer: The slope of the 
vertical line is undefined. 
The slope of the second line 
is 0. 



Thus, the slope of the horizontal 
line is zero, which makes sense be- 
cause a horizontal line neither goes 
uphill nor downhill. 



Division by zero is undefined. 
Hence, the slope of a vertical line is 
undefined. Again, this makes sense 
because as uphill lines get steeper 
and steeper, their slopes increase 
without bound. 



□ 



The Geometry of the Slope of a Line 

We begin our geometrical discussion of the slope of a line with an example, 
calculating the slope of a line passing through the points P(2,3) and Q(8, 8). 
Before we begin we'll first calculate the change in y and the change in x by sub- 
tracting the coordinates of point -P(2, 3) from the coordinates of point 



Slope 



Ay 

Ax 
8-3 



_ 5 

~~ 6 

Thus, the slope of the line through the points P(2, 3) and <5(8, 8) is 5/6. 

To use a geometric approach to finding the slope of the line, first draw the 
line through the points P(2,3) and (3(8,8) (see Figure 3.53). Next, draw a 
right triangle with sides parallel to the horizontal and vertical axes, using the 
points P(2,3) and Q(8,8) as vertices. As you move from point P to point R 
in Figure 3.53, note that the change in x is Ax = 6 (count the tick marks 1 ). 



1 When counting tick marks, make sure you know the amount each tick mark represents. 
For example, if each tick mark represents two units, and you count six tick marks when 
evaluating the change in x, then Ax = 12. 



3.3. RATES AND SLOPE 



193 



As you then move from point R to point Q, the change in y is Ay = 5 (count 
the tick marks). Thus, the slope is Ay/ Ax = 5/6, precisely what we got in the 
previous computation. 




Figure 3.53: Determining the slope 
of the line from the graph. 



10 
























R(6,S) 


Ax 


— 


b 




























Q 


(8 


8) 
























Ay = r, 










































































P 


(2 


3) 


























































-1_ 




1 
















in 



Figure 3.54: Determining the slope 
of the line from the graph. 



For contrast, in Figure 3.54, we started at the point P(2, 3), then moved upward 
5 units and right 6 units. However, the change in y is still Ay = 5 and the 
change in x is still Ax = 6 as we move from point P(2,3) to point Q(8,8). 
Hence, the slope is still Ay /Ax = 5/6. 

Consider a second example shown in Figure 3.55. Note that the line slants 
downhill, so we expect the slope to be a negative number. 



#(8,7) 



I 

in' 




























































(2 


7) 






















P 








































\( 


1 = 


- 


[ 






































(8 


3) 






















Q 




R( 


2- 


3) 




L 


\X 


= 


fi 


















































1 


p — 


1 
















1 







Ay = -4 



* x 



Figure 3.55: Determining the slope 
of the line from the graph. 



Figure 3.56: Determining the slope 
of the line from the graph. 



Rise over run. In 

Figure 3.54, we start at the 
point P(2,3), then "rise" 5 
units, then "run" 6 units to 
the right. For this reason, 
some like to think of the 
slope as "rise over run." 



In Figure 3.55, we've drawn a right triangle with sides parallel to the hor- In this case, the "rise" is 
izontal and vertical axes, using the points P(2,7) and Q(8,3) as vertices. As negative, while the "run" is 
you move from point P to point R in Figure 3.55, the change in y is Ay = —4 positive. 



194 



CHAPTER 3. INTRODUCTION TO GRAPHING 



(count the tick marks and note that your values of y are decreasing as you 
move from P to R). As you move from point R to point Q, the change is x is 
A2: = 6 (count the tick marks and note that your values of x are increasing as 
you move from R to Q). Thus, the slope is Ay/ Ax = —4/6, or —2/3. Note 
that the slope is negative, as anticipated. 

In Figure 3.56, we've drawn our triangle on the opposite side of the line. In 
this case, as you move from point P to point R in Figure 3.56, the change in x 
is Ax = 6 (count the tick marks and note that your values of x are increasing as 
you move from P to R). As you move from point R to point Q, the change is y 
is Ay = — 4 (count the tick marks and note that your values of y are decreasing 
as you move from J? to Q). Thus, the slope is still Ay/Ax = —4/6, or —2/3. 

We can verify our geometrical calculations of the slope by subtracting the 
coordinates of the point P (2, 7) from the point Q(8, 3). 

Q1 Ay 

Slope = — — 

Ax 

3-7 



You Try It! 



Sketch the line passing 
through the point (—4, 2) 
with slope —1/4. 



2 
~ ~3 
This agrees with the calculations made in Figures 3.55 and 3.56. 
Let's look at a final example. 



EXAMPLE 7. Sketch the line passing through the point (—2, 3) with slope 

-2/3. 

Solution: The slope is —2/3, so the line must go downhill. In Figure 3.57, 
we start at the point P{— 2,3), move right 3 units to the point _R(1,3), then 
move down 2 units to the point Q(l, 1). Draw the line through the points P 
and Q and you are done. 

In Figure 3.58, we take a different approach that results in the same line. 
Start at the point P'(— 2,3), move downward 4 units to the point R'{— 2, — 1), 
then right 6 units to the point Q'(4, — 1). Draw a line through the points P' 
and Q' and you are done. 

The triangle APQR in Figure 3.57 is similar to the triangle AP'Q'R' in 
Figure 3.58, so their sides are proportional. Consequently, the slope of the line 
through points P'(-2,3) and Q'(4,-l), 



Slope 



Ay 

Ax 
-4 

ir 

2 
~3 



3.3. RATES AND SLOPE 



195 











b 


















A 


r = 


3 


fl(l,3) 


P{- 


-2, 


3) 










2— 


























Q 


(1, 


1) 










5 


















5 






































































-5- 













Figure 3.57: Start at P(-2, 3), then 
move right 3 and down 2. The re- 
sulting line has slope —2/3. 













.y 
^4 












































P J (-2,3) 


































Ay 




-4 


































R' 


5 


















i 


u 


(-2,-1) 


Ax = ( 



































































^ 
























■ 


■ 













Figure 3.58: Starting at P'(-2,3) 
and moving down 4 and right 6 also 
yields a slope of —2/3. 



reduces to the slope of the line through the points P and Q in Figure 3.57. 



Answer: 









-5 J 

1 














-P(- 


Z O"! 












) - 


f 


>(0, 















<< 






















5 


















5 


































































- 


b 













□ 



A summary of facts about the slope of a line. We present a summary of 
facts learned in this section. 

1. The slope of a line is the rate at which the dependent variable is changing 
with respect to the independent variable. If y is the dependent variable 
and x is the independent variable, then the slope is 

Q1 Ay 

Slope =— , 

where Ay is the change in y (difference in y) and Aa; is the change in x 
(difference in x). 

2. If a line has positive slope, then the line slants uphill as you "sweep your 
eyes from left to right." If two lines have positive slope, then the line 
with the larger slope rises more quickly. 

3. If a line has negative slope, then the line slants downhill as you "sweep 
your eyes from left to right." If two lines have negative slope, then the 
line having the slope with the larger magnitude falls more quickly. 

4. Horizontal lines have slope zero. 

5. Vertical lines have undefined slope. 



196 CHAPTER 3. INTRODUCTION TO GRAPHING 



**. **• t* Exercises •** •** •** 



1. An object's initial velocity at time t = seconds is w = 10 meters per second. It then begins to 
pick up speed (accelerate) at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ). 

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. 
Include units with your labels. 

b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum 
of 5 additional points using the fact that the object is accelerating at a rate of 5 meters per 
second per second. 

c) Sketch the line representing the object's velocity versus time. 

d) Calculate the slope of the line. 

2. An object's initial velocity at time t = seconds is v = 40 meters per second. It then begins to 
lose speed at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ). 

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. 
Include units with your labels. 

b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum 
of 5 additional points using the fact that the object is losing speed at a rate of 5 meters per 
second per second. 

c) Sketch the line representing the object's velocity versus time. 

d) Calculate the slope of the line. 

3. David first sees his brother when the distance separating them is 90 feet. He begins to run toward 
his brother, decreasing the distance d between him and his brother at a constant rate of 10 feet 
per second (10 ft/s). 

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. 
Include units with your labels. 

b) Plot the point representing David's initial distance from his brother at time t = seconds. 
Then plot a minimum of 5 additional points using the fact that David's distance from his 
brother is decreasing at a constant rate of 10 feet per second (10 ft/s). 

c) Sketch the line representing David's distance from his brother versus time. 

d) Find the slope of the line. 



3.3. RATES AND SLOPE 



197 



4. David initially stands 20 feet from his brother when he sees his girl friend Mary in the distance. 
He begins to run away from his brother and towards Mary, increasing the distance d between him 
and his brother at a constant rate of 10 feet per second (10 ft/s). 

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. 
Include units with your labels. 

b) Plot the point representing David's initial distance from his brother at time t = seconds. 
Then plot a minimum of 5 additional points using the fact that David's distance from his 
brother is increasing at a constant rate of 10 feet per second (10 ft/s). 

c) Sketch the line representing David's distance from his brother versus time. 

d) Find the slope of the line. 



In Exercises 5-14, calculate the slope of the line passing through the points P and Q. Be sure to reduce 
your answer to lowest terms. 



5. P(9,0),Q(-9,15) 

6. P(19,-17), Q(-13,19) 

7. P(0,11), Q(16,-ll) 

8. P(-10,-8),Q(11,19) 

9. P(H,1), Q(-l,-l) 



10. P(16,-15), £(-11,12) 

11. P(-18,8), Q(3,-10) 

12. P(9,9), Q(-6,3) 

13. P(-18,10), Q(-9,7) 

14. P(-7,20), Q(7,8) 



In Exercises 15-18, use the right triangle provided to help determine the slope of the line. Be sure to 
pay good attention to the scale provided on each axis when counting boxes to determine the change in 
y and the change in x. 



15. 



16. 




*■ x 



10 



. 




























































\ 


































































































































\ 











*■ X 



20 



198 



CHAPTER 3. INTRODUCTION TO GRAPHING 



17. 



18. 











n 1 


/ 




















u 






































































































5 


















5 










































































-1 













\ 













r ' 






















o 






































































































10 


















10 


































































/ 








5- 















19. On one coordinate system, sketch each 
of the lines that pass through the fol- 
lowing pairs of points. Label each 
line with its slope, then explain the 
relationship between the slope found 
and the steepness of the line drawn. 

a) (0,0) and (1,1) 

b) (0,0) and (1,2) 

c) (0,0) and (1,3) 



20. On one coordinate system, sketch each 
of the lines that pass through the fol- 
lowing pairs of points. Label each 
line with its slope, then explain the 
relationship between the slope found 
and the steepness of the line drawn. 

a) (0,0) and (1,-1) 

b) (0,0) and (1,-2) 

c) (0,0) and (1,-3) 



In Exercises 21-30, setup a coordinate system on graph paper, then sketch the line passing through the 
point P with the slope m. 



21. P(-4,0), m = -3/7 

22. P(-3,0), m = -3/7 

23. P(-3,0), m = 3/7 

24. P(-3,0), m = 3/4 

25. P(-3,-3), m = 3/7 



26. P(-2,3), m= -3/5 

27. P(-4,3), m = -3/5 

28. P(-l,-3), m = 3/4 

29. P(-1,0), m= -3/4 

30. P(-3,3), m= -3/4 



3.3. RATES AND SLOPE 



199 



£*• 2*> £*■ Answers ■*$ -as >*$ 



v(m/s) 
50 

40 

30 

20 

10 





- . 
































































































































































a 


1 1 


fV 
















V 




-•-v 

















0123456789 10 



d(ft) 



t(s) 



100 

90 
80 
70 
GO 
50 
40 
30 
20 
10 




♦ (( 


u 


)U, 





































































































































































































0123456789 10 



t(s) 



5.-* 

6 



17. 



19. 



21. 



-m 3 = 3 




Ay = -3 



-54 



P(-4,0) 



A. 



-0(3,-3) 



r.-ii 



23. 



11. 



= 7" 



Ay = 



_Q(4,3) 



13. — 

3 



_P(-3,0). 



15.5 

2 



200 



CHAPTER 3. INTRODUCTION TO GRAPHING 



25. 



Ay 



V 

-54 



Ax = 1 



P(-3,-3). 

1-5-L-L 



4(4,0) 



29. 




0(3,-3) 



27. 



Ay = 











5 
3) 












i 


(- 


% 
































3 






\ 








(l 


" 
















Q 


5 




A; 


c = 


= 5 










5 








\ 






















s. 






























r. 













3.4. SLOPE-INTERCEPT FORM OF A LINE 



201 



3.4 Slope-Intercept Form of a Line 

We start with the definition of the y-intercept of a line. 

The y-intercept. The point (0, 6) where the graph of a line crosses the y-axis 
is called the y-intercept of the line. 

We will now generate the slope-intercept formula for a line having y-intercept 
(0, b) and Slope = m (see Figure 3.59). Let (x, y) be an arbitrary point on the 
line. 



Slope = m 




*■ x 



Figure 3.59: Line with y-intercept at (0, b) and Slope = m. 



Start with the fact that the slope of the line is the rate at which the dependent 
variable is changing with respect to the independent variable. 



Slope 



Ay 

Ax 



Slope formula. 



Substitute m for the slope. To determine both the change in y and the change 
in x, subtract the coordinates of the point (0, b) from the point (x, y). 



y-b 

x-0 
y-b 



Substitute m for the Slope. 
Ay = y — b and Ax = x — 0. 

Simplify. 



Clear fractions from the equation by multiplying both sides by the common 
denominator. 



y-b 



mx = y — b 



Multiply both sides by x. 
Cancel. 



202 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Solve for y. 



mx + b = y— b + b 
mx + b = y 



Add b to both sides. 
Simplify. 



Thus, the equation of the line is y = mx + b. 

The Slope-Intercept form of a line. The equation of the line having y- 
intercept (0, b) and slope m is: 

y = nix + b 

Because this form of a line depends on knowing the slope m and the intercept 
(0,6), this form is called the slope-intercept form of a line. 



You Try It! 



Sketch the graph of the line 
having equation 



y 



-x- 2. 



EXAMPLE 1. Sketch the graph of the line having equation y = —x + 1. 

5 

Solution: Compare the equation y = |a; + 1 with the slope-intercept form 
y = mx + b, and note that m = 3/5 and 6=1. This means that the slope is 
3/5 and the y-intercept is (0, 1). Start by plotting the y-intercept (0, 1), then 
move uward 3 units and right 5 units, arriving at the point (5, 4). Draw the line 
through the points (0, 1) and (5, 4), then label it with its equation y = |ir + 1. 





V 








3 






: ~y - 7 a 













^ 




Ax 


= 5 ^ 






U" 










Aw = 3 ; 


' "(5,4)- 




^_ 












.y "(0. 


1) r 


-1U 


^^ 


1U 




.^ 






S 




S 
































J — 10- J- 





x + l 



Figure 3.60: Hand-drawn graph of 




Figure 3.61: Select 6:ZStandard 
from the ZOOM menu to draw the 



graph of y 



ix + 1. 



y 



lx + 1. 



Check: Enter the equation y 



1 in Yl in the Y= menu as 3/5*X+l. 



Select 6:ZStandard from the ZOOM menu to produce the graph shown in 



3.4. SLOPE-INTERCEPT FORM OF A LINE 



203 



Figure 3.61. When we compare the calculator produced graph in Figure 3.61 to 
the hand drawn graph in Figure 3.60, they look a bit different. This difference 
is due to the fact that the calculator's viewing window is wider than it is tall. 
On the other hand, the grid in Figure 3.60 is perfectly square. Hence, it appears 
that the lines increase at different angles. 

The graphing calculator has a feature that will cure this distortion. Press 
the ZOOM button, the select 5:ZSquare to produce the image in Figure 3.63. 





V 








3 






: y-K x 













s 




Ax 


= 5 ^ 






U" 










Au = 3 ; 


;■* "(5,4)- 




<s_ 












.' (0, 


1) r 


-10 


^^ 


1U 




*7 






/ 




S 
































J — 10^ J- 






Figure 3.63: Select 5:ZSquare 
from the ZOOM menu to draw the 



graph of y 



ix + 1. 



Figure 3.62: Hand-drawn graph of 
V= tX+1. 



When we compare the calculator image in Figure 3.63 with the hand-drawn 
graph in Figure 3.62, we get a better match. 



Answer: 
4 

y 



10 4 


/ 






L 
































y 




> 




-10 5 


H 10 




\ 1 


10 -2 s ) 


^ /"Q «\- 




\ (3, — b) 


















LL— io- 





D 



EXAMPLE 2. Sketch the line with y-intercept (0, 2) and slope -7/3. Label 
the line with the slope-intercept form of its equation. 

Solution: Plot the y-intercept (0,2). Now use the slope —7/3. Start at (0,2), 
then move down seven units, followed by a three unit move to the right to 
the point (3,-5). Draw the line through the points (0,2) and (3,-5). (see 
Figure 3.64). 

Next, the y-intercept is (0,2), so b = 2. Further, the slope is —7/3, so 
m = —7/3. Substitute these numbers into the slope-intercept form of the line. 



y 



y 



mx 

7 



Slope-intercept form. 
Substitute: —7/3 for m, 2 for b. 



Therefore, the slope-intercept form of the line is y 
with this equation. 



2. Label the line 



You Try It! 



Sketch the line with 
y-intercept (0, —3) and slope 
5/2. Label the line with the 
slope-intercept form of its 
equation. 



204 



CHAPTER 3. INTRODUCTION TO GRAPHING 



y = - -x + 2 y 



yfy -J^-W- 




L 




5 




L_ 




\ 




A 




c 


-(0,2). _ 


i 






v 




5 




u in 


-10 Aw = -7 


A -H 




A 




■ V-(3,-5). 








1 


A 


x = 3 




L_ 




V 


J — 10- 


A 



Figure 3.64: Hand-drawn graph of y = — Zx + 2. 



Check: To graph y = — Zx + 2, enter -7/3*X+2 in Yl in the Y= menu. 
Select 6:ZStandard from the ZOOM menu, followed by 5:ZSquare from the 
ZOOM menu to produce the graph shown in Figure 3.66. 



Answer: 



llO 



y y = 2 x ~ 3 



-10: 



(o. 



m 



I 



2,2 



10 



= --x + 2 y 




^ 1U 




L 




5 




L_ 




\ 




A 




r 


-(0,2). 


i 




h 






\" 




u in 


-10 Aw = -7 


A -n 




A 




■ V-(3,-5). 








*\ 


A 


x = 3 




L_ 




V 


J — 10- 


A 



Figure 3.65: Hand-drawn graph of 

y 



■lx + 2. 




Figure 3.66: Select 6:ZStandard 
from the ZOOM menu, followed by 
5:ZSquare from the ZOOM menu 
to produce the graph of the equa- 
tion y = —ix + 2. 



This provides a good match of the hand-drawn graph in Figure 3.65 and our 
graphing calculator result in Figure 3.66. 



□ 



3.4. SLOPE-INTERCEPT FORM OF A LINE 



205 



You Try It! 



EXAMPLE 3. Use the graph of the line in the following figure to find the Use the graph of the line in 
equation of the line. the figure below to find the 

equation of the line. 



V 



















il 


H 




































































































































































































































































































































/ 








































/ 


















-If 




















/ 


















10 


















/ 




































/ 


' 






































/ 






































/ 






































/ 








































/ 






































/ 






































/ 






































/ 


' 








If 


)■ 


■ 





















I 








% 




v.. 




N ^ r 




% 




N v 




N S 








■s 




^^ 


T 


> *. T 


-10 


N 1Q 






























U-— io- 





Solution: Note that the y-intercept of the line is (0,-1) (see Figure 3.67). 
Next, we try to locate a point on the line that passes directly through a lattice 
point, a point where a vertical and horizontal grid line intersect. In Figure 3.67, 
we chose the point (5, 6). Now, starting at the ^-intercept (0, 1), we move up 7 
units, then to the right 5 units. Hence, the slope is m = Ay/Aa;, or m = 7/5. 



y 



We could also subtract the 
coordinates of point (0,-1) 
from the coordinates of point 
(5, 6) to determine the slope. 



(-1) 



5-0 

















10 J 


■ 








/ 




























/ 
























Air 


c 


1 


/ 
























/ 






























r , 




























/ 


( 


5,0! 




















/ 




















L 


\y = 


V 


/ 


/ 




























/_ 






























/ 














If 
















t - 












1( 














) 






























-(0,-j 














































































































































































































10n 


1 — — 















Figure 3.67: The line has y-intercept (0, —1) and slope 7/5. 



206 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Next, state the slope-intercept form, the substitute 7/5 for m and —1 for b. 



y = mx + b 

V=-x + {-\) 
5 

Thus, the equation of the line is y 



Slope-intercept form. 
Substitute: 7/5 for to, — 1 for 

= lx-l. 



Check: This is an excellent situation to run a check on your graphing calcu- 
lator. 



Ploti Plots Plots 
■^Y£ = 

\Yh = 
■^Ye = 

--W? = 




Figure 3.68: Enter y 
the Y= menu. 



1 in 



Figure 3.69: Select 6:ZStandard 
followed by 5:ZSquare (both from 
the ZOOM menu) to produce this 
graph. 



Answer: y = x 

y 5 



When we compare the result in Figure 3.69 with the original hand-drawn graph 
(see Figure 3.67), we're confident we have a good match. 

□ 



You Try It! 



A swimmer is 50 feet from 
the beach, and then begins 
swimming away from the 
beach at a constant rate of 
1.5 feet per second (1.5 ft/s) 
Express the distance d 
between the swimmer and 
the beach in terms of the 
time t. 



Applications 

Let's look at a linear application. 



EXAMPLE 4. Jason spots his brother Tim talking with friends at the 
library, located 300 feet away. He begins walking towards his brother at a 
constant rate of 2 feet per second (2 ft/s). 

a) Express the distance d between Jason and his brother Tim in terms of the 
time t. 

b) At what time after Jason begins walking towards Tim are the brothers 200 
feet apart? 

Solution: Because the distance between Jason and his brother is decreasing 
at a constant rate, the graph of the distance versus time is a line. Let's begin 



3.4. SLOPE-INTERCEPT FORM OF A LINE 



207 



by making a rough sketch of the line. In Figure 3.70, note that we've labeled 
what are normally the x- and y-axes with the time t and distance d, and we've 
included the units with our labels. 

d(ft) 



. 


. 




















(0,300) 




































200 






































100 






































n 





















t(8) 

50 100 150 200 

Figure 3.70: A plot of the distance d separating the brothers versus time t. 



Let t = seconds be the time that Jason begins walking towards his brother 
Tim. At time t = 0, the initial distance between the brothers is 300 feet. 
This puts the <i-intercept (normally the y-intercept) at the point (0, 300) (see 
Figure 3.70). 

Because Jason is walking toward his brother, the distance between the 
brothers decreases at a constant rate of 2 feet per second. This means the 
line must slant downhill, making the slope negative, so m = — 2 ft/s. We can 
construct an accurate plot of distance versus time by starting at the point 
(0, 300), then descending Ad = —300, then moving to the right At = 150. This 
makes the slope Ad/ At = -300/150 = -2 (see Figure 3.70). Note that the 
slope is the rate at which the distance d between the brothers is changing with 
respect to time t. 

Finally, the equation of the line is y = mx + b, where m is the slope of the 
line and b is the y-coordinate (in this case the (i-coordinate) of the point where 
the graph crosses the vertical axis. Thus, substitute —2 for m, and 300 for b 
in the slope-intercept form of the line. 



y = mx + b 
y = -2x + 300 



Slope-intercept form. 
Substitute: —2 for m, 300 for b. 



One problem remains. The equation y = —2x + 300 gives us y in terms of x. 
The question required that we express the distance d in terms of the time t. 
So, to finish the solution, replace y with d and x with t (check the axes labels 



208 CHAPTER 3. INTRODUCTION TO GRAPHING 

in Figure 3.70) to obtain a solution for part (a): 

d = -2i + 300 

Now that our equation expresses the distance between the brothers in terms of 
time, let's answer part (b), "At what time after Jason begins walking towards 
Tim are the brothers 200 feet apart?" To find this time, substitute 200 for d 
in the equation d = —2t + 300, then solve for t. 

d = —2t + 300 Distance equation. 

200 = -2t + 300 Substitute 200 for d. 

Solve this last equation for the time t. 

200 - 300 = -2t + 300 - 300 Subtract 300 from both sides. 

— 100 = — It Simplify both sides. 



-100 -2t 



Divide both sides by —2. 



-2 -2 

50 = t Simplify both sides. 

Thus, it takes Jason 50 seconds to close the distance between the brothers to 
Answer: d = 1.5* + 50 200 feet. 

□ 



3.4. SLOPE-INTERCEPT FORM OF A LINE 



209 



ti. ;». ;». 



Exercises 



■*j ■*-: ■*; 



In Exercises 1-6, setup a coordinate system on graph paper, then sketch the line having the given 
equation. Label the line with its equation. 



1. y = |x - 6 

2. j/=fz-l 

3.y = -^x + 4 



4- y = jx 

5. y = -^x + 4 

6. y = ~lx + 7 



In Exercises 7-12, sketch the line with given y-intercept slope. Label the line with the slope-intercept 
form of its equation. 



7. (0,-7), 9/5 

8. (0,7), -4/5 

9. (0,-1), 6/7 



10. (0,1), -7/5 

11. (0,-6), 9/7 

12. (0,-5), 7/5 



In Exercises 13-20, determine the equation of the given line in slope-intercept form. 
13. 14. 



ti 



6 \ 6 

— -'6-1—1 — t- 




210 



CHAPTER 3. INTRODUCTION TO GRAPHING 



15. 



18. 



"Tvl 6 f 

6 \ 6 

— 6-1— I -^ 













6 J 






































































































































fi 






















fi 
















































































/ 




















/ 




fi 















16. 



19. 













I 
(■ ' 


/ 










































































































































































fi 






















f 




















/ 




































































/ 






















/ 


s 






fi 



























fi' 


















































































































































fi 






















fi 




















































































































fi 

















17. 



20. 













0' 






































































































































fi 






















fi 
















\ 


s 


























































































fi 

























1 


/ 










































































































































































fi 






















f 






















































































































fi 

















3.4. SLOPE-INTERCEPT FORM OF A LINE 



211 



21. Assume that the relationship between an 
object's velocity and its time is linear. At 
t = seconds, the object's initial velocity 
is 20 m/s. It then begins to speed up at a 
constant rate of 5 meters per second per 
second. 

a) Set up a coordinate system, placing the 
time t on the horizontal axis and the ve- 
locity v on the vertical axis. Label and 
scale each axis. Include units in your la- 
bels. 

b) Use the initial velocity and the rate at 
which the object's velocity is increasing to 
draw the line representing the object's ve- 
locity at time t. Use the slope-intercept 
form to determine the equation of the line. 

c) Replace x and y in the equation found in 
part (b) with t and v, respectively. Use 
the result to determine the velocity of the 
object after 14 seconds. 

22. Assume that the relationship between an 
object's velocity and its time is linear. At 
t = seconds, the object's initial velocity 
is 80 m/s. It then begins to lose speed at 
a constant rate of 4 meters per second per 
second. 

a) Set up a coordinate system, placing the 
time t on the horizontal axis and the ve- 
locity v on the vertical axis. Label and 
scale each axis. Include units in your la- 
bels. 

b) Use the initial velocity and the rate at 
which the object's velocity is increasing to 
draw the line representing the object's ve- 
locity at time t. Use the slope-intercept 
form to determine the equation of the line. 

c) Replace x and y in the equation found in 
part (b) with t and v, respectively. Use 
the result to determine the velocity of the 
object after 13 seconds. 

23. A water tank initially (at time t = min) 
contains 100 gallons of water. A pipe is 
opened and water pours into the tank at 



a constant rate of 25 gallons per minute. 
Assume that the relation between the vol- 
ume V of water in the tank and time t is 
linear. 

a) Set up a coordinate system, placing the 
time t on the horizontal axis and the vol- 
ume of water V on the vertical axis. Label 
and scale each axis. Include units in your 
labels. 

b) Use the initial volume of water in the tank 
and the rate at which the volume of water 
is increasing to draw the line representing 
the volume V of water in the tank at time 
t. Use the slope- intercept form to deter- 
mine the equation of the line. 

c) Replace x and y in the equation found in 
part (b) with t and V, respectively. Use 
the result to predict how much time must 
pass until the volume of water in the tank 
reaches 400 gallons. 

24. A water tank initially (at time t = min) 
contains 800 gallons of water. A spigot 
is opened at the bottom of the tank and 
water pours out at a constant rate of 40 
gallons per minute. Assume that the re- 
lation between the volume V of water in 
the tank and time t is linear. 

a) Set up a coordinate system, placing the 
time t on the horizontal axis and the vol- 
ume of water V on the vertical axis. Label 
and scale each axis. Include units in your 
labels. 

b) Use the initial volume of water in the tank 
and the rate at which the volume of water 
is decreasing to draw the line representing 
the volume V of water in the tank at time 
t. Use the slope-intercept form to deter- 
mine the equation of the line. 

c) Replace x and y in the equation found in 
part (b) with t and V, respectively. Use 
the result to predict how much time must 
pass until the water tank is empty. 



212 



CHAPTER 3. INTRODUCTION TO GRAPHING 



s*- $*> £»■ Answers •*$ •*& ■** 



-10 



Ay 



10 



Ax = 



«6 



/ 



/ 



/ 



/, 



2/= la: -6 



0(5,3) 

■ f x 



P(0,-6) 





10 



3. 





) 


J M- 




r_ 




3_ 






T~t/r\ a \ 


) 


"(0,4); 












5 








\ T 




\ T '*-' 


lAy = —11" 


\ 10 




i -4-a 




3 zq 








t 0(4,-7) 




Jh 






/ 


ix — 4 


- 10- 


' jr - 11 



x + 4 




y = ji- 1 



0(7,5) 



11. 



-10 



10 



Ay = 



y = fa; -6 



0(7,3) 

■ t x 



P(0,-6) 







10 



\ 



10 



1 



-lAy = -11 



-10 



P(0, 



\ 



4) 



Ax = 7- 
"111 



10 



7. 



-10 



:Ay ; 



y 
10+ 



-id 



Ax = 



I 



/ 



I 



P(0, 



0(7, -7) 

y = -±fx + 4 



9 r7 

y = 5^-7 



/ 



0(5, J8)s 
10 



-7) 



13. y = --x 
y 3 



15. y = —-x + 1 



17. j/= --x 



19. y = -x - 4 
5 



21. c) 90 m/s 



23. c) 12 min 



3.5. POINT-SLOPE FORM OF A LINE 



213 



3.5 Point-Slope Form of a Line 

In the previous section we learned that if we are provided with the slope of a 
line and its y-intercept, then the equation of the line is y = mx + 6, where m 
is the slope of the line and b is the y-coordinate of the line's y-intercept. 

However, suppose that the y-intercept is unknown? Instead, suppose that 
we are given a point (xo 1 2/o) on the line and we're also told that the slope of 
the line is m (see Figure 3.71). 




* x 



Slope = m 



Figure 3.71: A line through the point (xo,yo) with slope m. 



Let (x,y) be an arbitrary point on the line, then use the points (xo,yo) and 
(x, y) to calculate the slope of the line. 



Slope 



Ay 

Ax 

y-yo 

X — Xq 



The slope formula. 

Substitute m for the slope. Use (x, y) 
and (xo,j/o) to calculate the difference 
in y and the difference in x. 



Clear the fractions from the equation by multiplying both sides by the common 
denominator. 



m(x — Xo) 



y-yo 



_x — Xo 
m(x- x ) =y-yo 



(x — xo) Multiply both sides by x — xo- 
Cancel. 



Thus, the equation of the line is y — yo = m(x — Xq) 



214 



CHAPTER 3. INTRODUCTION TO GRAPHING 



You Try It! 



Draw the line passing 
through the point (1,-2) 
that has slope 3/2, then 
label it with its equation. 



Answer: 











- ' 


/ 














































































































5 


















5 
































( 




r 


^ 














\ 


L J "J 










K 







y + 2=-( X -l) 



The Point 


-Slope 


form of . 


i line. 


The equation 


of the line with slope m 


that 


passes through the 


point (xq 


,Vo) is: 














y-yo 


= m(x — xq) 







EXAMPLE 1. Draw the line passing through the point (—3,-1) that has 
slope 3/5, then label it with its equation. 

Solution: Plot the point (—3,-1), then move 3 units up and 5 units to the 
right (see Figure 3.72). To find the equation, substitute (—3,-1) for (xo,yo) 
and 3/5 for m in the point-slope form of the line. 



y-y = m(x -Xq) 

V-(-l) = f(*-(-3)) 



Point-slope form. 

Substitute: 3/5 for m, —3 
for xq, and —1 for j/q- 



Simplifying, the equation of the line is y + 1 = |(a; + 3). 



Ay 











- ' 
























o 












y + l = 


= |( !B + 3 ) 








A: 


c = 


= 5 








/ 








































= 3 






/ 














5 


















\ 










(-3.-1) 


































































x, 


■ 













Figure 3.72: The line passing through (—3, —1) with slope 3/5. 



□ 

At this point, you may be asking the question "When should I use the slop- 
intercept form and when should I use the point-slope form?" Here is a good 
tip. 



Tip. Which form should I use? The form you should select depends upon 
the information given. 



3.5. POINT-SLOPE FORM OF A LINE 



215 



1. If you are given the y-intercept and the slope, use y = mx + b. 

2. If you are given a point and the slope, use y — yo = m(x — xq). 



You Try It! 



EXAMPLE 2. Find the equation of the line passing through the points Find the equation of the line 

P(— 1, 2) and Q(3, —4). passing through the points 

P(-2,l) andQ(4,-l). 
Solution: First, plot the points P(— 1, 2) and <5(3, —4) and draw a line through 

them (see Figure 3.73). 




0(3,-4) 



Figure 3.73: The line passing through P (— 1, 2) and Q(3, —4). 

Next, let's calculate the slope of the line by subtracting the coordinates of 
the point P(— 1, 2) from the coordinates of point Q(3, —4). 

Slope = — — 

Ax 

-4-2 
~3-(-l) 
-6 



3 

~~2 

Thus, the slope is —3/2. 

Next, use the point-slope form y — yo = m(x—xo) to determine the equation 
of the line. It's clear that we should substitute —3/2 for m. But which of the 
two points should we use? If we use the point P[— 1, 2) for (xq, yo), we get the 
answer on the left, but if we use the point Q(3, — 4) for (xo,yo), we get the 



216 



CHAPTER 3. INTRODUCTION TO GRAPHING 



answer on the right. 



y 



-jOs-C- 1 )) 



y- (-4) = --(*- 3) 



At first glance, these answers do not look the same, but let's examine them a 
bit more closely, solving for y to put each in slope-intercept form. Let's start 
with the equation on the left. 



Note that the form 
y = — \x + \, when 
compared with the general 
slope-intercept form 
y = mx + b, indicates that 
the y-intercept is (0, 1/2). 
Examine Figure 3.73. Does 
it appear that the 
y-intercept is (0, 1/2)? 



1 1 

Answer: y = x H — 

y 3 3 



y-2 

y-2 

2/-2 
y-2 + 2 

y 



- 5 (* -M)) 



(x + l) 
3 



3 

2 

3 

—x 

2 2 

3 3 

2 X ~ 2 

3 3 

— x 

2 2 



3 1 

--x+ - 
2 2 



Using m = —3/2 and (xq, j/o) = (— 1, 2). 

Simplify. 

Distribute -3/2. 

Add 2 to both sides. 

On the left, simplify. On the right, 
make equivalent fractions with a 
common denominator. 

Simplify. 



Let's put the second equation in slope-intercept form. 



y-(-4) = 


-5(*-3) 




Using m = —3/2 and (xo,yo) = (3, — 


y + 4 = 


-§(*-3) 




Simplify. 


y + A = 


3 9 

--ar+ - 
2 2 




Distribute -3/2. 


y + 4- 4 = 


3 9 

— jc H 

2 2 


4 


Subtract 4 from both sides. 


2/ = 


3 9 

— x H 


8 


On the left, simplify. On the right, 




2 2 


2 


make equivalent fractions with a 
common denominator. 


y = 


3 1 

— x H — 
2 2 




Simplify. 



Thus, both equations simplify to the same answer, y = — Ice + i. This means 
that the equations y — 2 = — |(x — (—1)) and y — (—4) = — I(cc — 3), though 
they look different, are the same. 

□ 



3.5. POINT-SLOPE FORM OF A LINE 



217 



Example 2 gives rise to the following tip. 

Tip. When finding the equation of a line through two points P and Q, you 
may substitute either point P or Q for (xq, J/o) m the point-slope form y — yo = 
m(x — xq). The results look different, but they are both equations of the same 
line. 



Parallel Lines 

Recall that slope is a number that measures the steepness of the line. If two 
lines are parallel (never intersect), they have the same steepness. 

Parallel lines. If two lines are parallel, they have the same slope. 



You Try It! 



EXAMPLE 3. Sketch the line y 



2, then sketch the line passing 



through the point (—1,1) that is parallel to the line y 
equation of this parallel line. 



2. Find the 



Solution: Note that y = |x — 2 is in slope-intercept form y = mx + b. Hence, 
its slope is 3/4 and its ^-intercept is (0, —2). Plot the y-intercept (0, —2), move 
up 3 units, right 4 units, then draw the line (see Figure 3.74). 



Find the equation of the line 
which passes through the 
point (2, —3) and is parallel 
to the line 



y 



-x + l. 




*■ X 




> X 



Figure 3.74: The line y = jx 



Figure 3.75: Adding a line parallel 
toy= fa: -2. 



The second line must be parallel to the first, so it must have the same slope 
3/4. Plot the point (— 1, 1), move up 3 units, right 4 units, then draw the line 
(see the red line in Figure 3.75). 



218 



CHAPTER 3. INTRODUCTION TO GRAPHING 



To find the equation of the parallel red line in Figure 3.75, use the point- 
slope form, substitute 3/4 for to, then (—1, 1) for (%o,yo). That is, substitute 
— 1 for xq and 1 for y^. 



y-yo = m(x-x Q ) 
y-l = !(*-(-!)) 



y-l=-(x + l) 



Point-slope form. 

Substitute: 3/4 for to, — 1 for xo, 
and 1 for g/o- 

Simplify. 



Plotl PlotE Plots 
WiB3^4*K-2 

Wj = 
\Yh= 

We = 



Figure 3.76: Enter equations 
of parallel lines. 



WINDOW 
Xmin= _ 5 
Xmax=5 
Xscl=l 
Vmin= _ 5 
Vmax=5 
Vscl=l 



Figure 3.77: Adjust the 
WINDOW parameters as 
shown. 



Check: In this example, we were not required to solve for y, so we can save 
ourselves some checking work by writing the equation 



y — 1 = — (x + 1) in the form 



y = -(x + l) + l 



by adding 1 to both sides of the first equation. Next, enter each equation as 
shown in Figure 3.76, then change the WINDOW setting as shown in Figure 3.77. 
Next, press the GRAPH button, the select 5:ZSquare to produce the image 
in Figure 3.79. 




.....jf 


•// 







Figure 3.79: Press the GRAPH 
button then select 5:ZSquare to 
produce this image. 



Figure 3.78: Hand-drawn parallel 
lines. 



Answer: y = —x 
y 2 



Note the close correlation of the calculator lines in Figure 3.79 to the hand- 
drawn lines in Figure 3.78. This gives us confidence that we've captured the 
correct answer. 

□ 



3.5. POINT-SLOPE FORM OF A LINE 



219 



Perpendicular Lines 

Two lines are perpendicular if they meet and form a right angle (90 degrees) . 
For example, the lines £i and £2 in Figure 3.80 are perpendicular, but the 
lines C\ and £2 in Figure 3.81 are not perpendicular. 





Figure 3.80: The lines £1 and £2 
are perpendicular. They meet and 
form a right angle (90 degrees). 



Figure 3.81: The lines C\ and £2 
are not perpendicular. They do 
not form a right angle (90 degrees) . 



Before continuing, we need to establish a relation between the slopes of 
two perpendicular lines. So, consider the perpendicular lines £1 and £2 in 
Figure 3.82. 




Figure 3.82: Perpendicular lines £1 and £2 



Things to note: 

1. If we were to rotate line £1 ninety degrees counter-clockwise, then C\ 
would align with the line £2, as would the right triangles revealing their 
slopes. 



Ay mi 

2. £1 has slope = — 

F Ax 1 



mi. 



220 



CHAPTER 3. INTRODUCTION TO GRAPHING 



3. £2 has slope 



Ay 

Ax 



-mi 



1 

mi 



Slopes of perpendicular lines. If C\ and £2 are perpendicular lines and C\ 
has slope mi, the £2 has slope — 1/mi. That is, the slope of £2 is the negative 
reciprocal of the slope of C\. 

Examples: To find the slope of a perpendicular line, invert the slope of the 
first line and negate. 

• If the slope of £1 is 2, then the slope of the perpendicular line £2 is — 1/2. 

• If the slope of £1 is —3/4, then the slope of the perpendicular line £2 is 

4/3. 



You Try It! 



Find the equation of the line 
that passes through the 
point (—3, 1) and is 
perpendicular to the line 



y 



1. 



EXAMPLE 4. Sketch the line y = — |cc — 3, then sketch the line through 
(2, 1) that is perpendicular to the line y = — fa; — 3. Find the equation of this 
perpendicular line. 

Solution: Note that y = — fa; — 3 is in slope-intercept form y = mx + b. Hence, 
its slope is —2/3 and its y-intercept is (0,-3). Plot the y-intercept (0,-3), 
move right 3 units, down two units, then draw the line (see Figure 3.83). 




Figure 3.83: The line y = — fa; — 3. 













/ 
























1 


\x 




; / 


































A 


y = 


3 
































S 














(2 


1) 






5 








































































/ 






















/ 


-5- 















Figure 3.84: Adding a line perpen- 
dicular to y = — \x — 3. 



Because the line y = -»a:-3 has slope —2/3, the slope of the line perpendicular 
to this line will be the negative reciprocal of —2/3, namely 3/2. Thus, to draw 
the perpendicular line, start at the given point (2, 1), move up 3 units, right 2 
units, then draw the line (see Figure 3.84). 



3.5. POINT-SLOPE FORM OF A LINE 



221 



To find the equation of the perpendicular line in Figure 3.84, use the point- 
slope form, substitute 3/2 for to, then (2, 1) for (xo,yo)- That is, substitute 2 
for xq, then 1 for y$. 



y-yo 
y-i 



m(x — Xq) 



Point-slope form. 

Substitute: 3/2 for m, 2 for xq, 
and 1 for y$. 



Check: In this example, we were not required to solve for y, so we can save 
ourselves some checking work by writing the equation 



y 



3, 
1 = -(x 

2 V 



in the form 



y 



-{x-2) + \ 



by adding 1 to both sides of the first equation. Next, enter each equation 
as shown in Figure 3.85, then select 6:ZStandard to produce the image in 
Figure 3.86. 



Ploti Plots Plots 
\ViB-2^3*X~3 
W2B3/-2*<X-2>+i 

\Vi = 
--W? = 




Figure 3.85: Enter equations of per- 
pendicular lines. 



Figure 3.86: 6:ZStandard 

duces two lines that do not 
perpendicular. 



pro- 
look 



Note that the lines in Figure 3.86 do not appear to be perpendicular. Have 
we done something wrong? The answer is no! The fact that the calculator's 
viewing screen is wider than it is tall is distorting the angle at which the lines 
meet. 

To make the calculator result match the result in Figure 3.84, change the 
window settings as shown in Figure 3.87, then select 5:ZSquare from the ZOOM 
menu to produce the image in Figure 3.88. Note the close correlation of the 
calculator lines in Figure 3.88 to the hand-drawn lines in Figure 3.84. This 
gives us confidence that we've captured the correct answer. 



Answer: y = 2x + 7 



□ 



222 



CHAPTER 3. INTRODUCTION TO GRAPHING 



WINDOW 
Xnun= - 5 
Xnax=5 
Xscl=l 
Vnun= - 5 
Vroax=5 
Vscl=l 

■i-Xres=l 




Figure 3.87: Change the WIN- 
DOW parameters as shown. 



Figure 3.88: 5:ZSquare produces 
two lines that do look perpendicu- 
lar. 



You Try It! 



Find an equation that 
expresses the Fahrenheit 
temperature in terms of the 
Celsius temperature. Use the 
result to find the Fahrenheit 
temperature when the 
Celsius temperature is 25° C. 



Applications 

Let's look at a real- world application of lines. 



EXAMPLE 5. Water freezes at 32° F (Fahrenheit) and at 0° C (Celsius). 
Water boils at 212° F and at 100° C. If the relationship is linear, find an equa- 
tion that expresses the Celsius temperature in terms of the Fahrenheit tem- 
perature. Use the result to find the Celsius temperature when the Fahrenheit 
temperature is 113° F. 

Solution: In this example, the Celsius temperature depends on the Fahren- 
heit temperature. This makes the Celsius temperature the dependent variable 
and it gets placed on the vertical axis. This Fahrenheit temperature is the 
independent variable, so it gets placed on the horizontal axis (see Figure 3.89). 

Next, water freezes at 32° F and 0° C, giving us the point (F, C) = (32, 0). 
Secondly, water boils at 212° F and 100° C, giving us the point (F, C) = 
(212,100). Note how we've scaled the axes so that each of these points fit 
on the coordinate system. Finally, assuming a linear relationship between the 
Celsius and Fahrenheit temperatures, draw a line through these two points (see 
Figure 3.89). 

Calculate the slope of the line. 



Slope formula. 





AC 




m = 


AF 






100- 


-0 




212- 

100 


-32 


m = 


180 
5 




m = 


9 





Use the points (32,0) and (212, 100). 
to compute the difference in C and F. 

Simplify. 
Reduce. 



You may either use (32,0) or (212,100) in the point-slope form. The point 
(32,0) has smaller numbers, so it seems easier to substitute (cco,j/o) = (32,0) 



3.5. POINT-SLOPE FORM OF A LINE 



223 



C 



120 



100 



SO 



GO 



40 



20 



-20 



-40 



(32,0) 



(212,100) 



40 60 80 100 120 140 160 180 200 220 240 



Figure 3.89: The linear relationship between Celsius and Fahrenheit tempera- 
ture. 



and m = 5/9 into the point-slope form y — yo = rn(x — Xq). 



1 - 2/o = rn(x - Xq) 


Point-slope form. 


y-0=^(x-32) 


Substitute: 5/9 for m, 32 for xo, 
and for yo- 


V=\(?- 32) 


Simplify. 



However, our vertical and horizontal axes are labeled C and F (see Figure 3.89) 
respectively, so we must replace y with C and x with F to obtain an equation 
expressing the Celsius temperature C in terms of the Fahrenheit temperature 
F. 

C=^(F-32) (3.1) 

Finally, to find the Celsius temperature when the Fahrenheit temperature is 
113° F, substitute 113 for F in equation (3.1). 



C 
C 

c 

c 



9 
5 

9 

5 

9 

45 



(F-32) 
(113-32) 



(81) 



Equation (3.1). 

Substitute: 113 for F. 

Subtract. 
Multiply. 



Therefore, if the Fahrenheit temperature is 113° F, then the Celsius tempera- 
ture is 45° C. 



Answer: 77° F 



□ 



224 CHAPTER 3. INTRODUCTION TO GRAPHING 



f> t* t* Exercises «•* -*$ •** 

In Exercises 1-6, set up a coordinate system on a sheet of graph paper, then sketch the line through 
the given point with the given slope. Label the line with its equation i point-slope form. 

1. m = -5/7, P(-3,4) 4. m = 5/6, P(-3,-l) 

2. m = 3/4, P(-2, -4) 5. m = 5/3, P(-l, -4) 

3. m = -4/5, P(-2, 4) 6. m = -3/8, P(-4, 0) 

In Exercises 7-12, set up a coordinate system on a sheet of graph paper, then sketch the line through 
the given point with the given slope. Label the line with the point-slope form of its equation. 

7. P(-4, 0) and Q(4, 3) 10. P(-3, 4) and Q(2, 0) 

8. P(-2,4) and 0(2,-1) 11. P(-3, 1) and 0(4,-4) 

9. P(-3, -3) and 0(2, 0) 12. P(-l, 0) and 0(4, 3) 



In Exercises 13-18, on a sheet of graph paper, sketch the given line. Plot the point P and draw a line 
through P that is parallel to the first line. Label this second line with its equation in point-slope form. 

13.y = -fa; + l, P(-3,2) 

14. y = -|x, P(-4,0) 

15. y=jk P(-4,0) 



In Exercises 19-24, on a sheet of graph paper, sketch the line passing through the points P and Q. Plot 
the point R and draw a line through R that is perpendicular to the line passing through the points P 
and 0- Label this second line with its equation in point-slope form. 

19. P(-2, 0), Q(2, -3), and P(-l, 0) 22. P(-4, -4), Q(-l, 3), and P(-4, 2) 

20. P(-l,3), 0(2,-2), and P(-1,0) 23. P(-2,3), 0(1,-1), and #(-3,-4) 

21. P(-2,-4), 0(1,4), and P(-4,-l) 24. P(-4,4), 0(1,-4), and #(-4,-3) 



16. 


y = fa;- 1, 


P(-2,-2) 


17. 


y= jx + i, 


P(-3,0) 


18. 


y = fa; -4, 


P(-2,-2) 



3.5. POINT-SLOPE FORM OF A LINE 



225 



25. Assume that the relationship between an 
object's velocity and its time is linear. At 
3 seconds, the object's velocity is 50 ft/s. 
At 14 seconds, the object's velocity is 30 

ft/s. 

a) Set up a coordinate system, placing 
the time t on the horizontal axis and 
the velocity v on the vertical axis. La- 
bel and scale each axis. Include units 
in your labels. 

b) Plot the points determined by the 
given data and draw a line through 
them. Use the point-slope form of the 
line to determine the equation of the 
line. 

c) Replace x and y in the equation found 
in part (b) with t and v, respectively, 
then solve the resulting equation for 

v. 

d) Use the result of part (c) to determine 
the velocity of the object after 6 sec- 
onds. 



26. Water freezes at approximately 32° F and 
273° K, where F represents the tempera- 
ture measured on the Fahrenheit scale and 
K represents the temperature measured 
on the Kelvin scale. Water boils at ap- 
proximately 212° F and 373° K. Assume 
that the relation between the Fahrenheit 
and Kelvin temperatures is linear. 

a) Set up a coordinate system, placing 
the Kelvin temperature K on the hor- 
izontal axis and the Fahrenheit tem- 
perature F on the vertical axis. Label 
and scale each axis. Include units in 
your labels. 

b) Plot the points determined by the 
given data and draw a line through 
them. Use the point-slope form of the 
line to determine the equation of the 
line. 

c) Replace x and y in the equation found 
in part (b) with K and F, respec- 
tively, then solve the resulting equa- 
tion for F. 

d) Use the result of part (c) to deter- 
mine the Fahrenheit temperature of 
the object if the Kelvin temperature 
is 212° K. 



;:»■;■*■;*. 



Answers 



■*s ■*■> •*; 



Ay = 




0(4,-1) 
6 



•4 = -f(a: + 3) 




0(3,0) 

y x 



y-4 



Ux + 2) 



226 



CHAPTER 3. INTRODUCTION TO GRAPHING 



5. 



11. 











y 




























/ 


.y 


















/ 


















/ 














\x = 


rj 




































7 nfo iv 


(j 










T 




' \^j 


5 




Ay- ' 


z 




















/ 




















t 


















/ 


Pi i a\ 










y 


^ 




; 







W + 4=|(x + l) 



-P(-4,0)- 



y=§(x + 4) 



Q(4,3) 
> a; 



Alternate answer: y — 3 = | (x — 4) 




•f(z + 3) 



Alternate answer: y + 4 = — = (x — 4) 



13. 



,, = -§35 + 1 







\ 




— 


; •■ 




















V 
























P( ^ 9^ 
















*A 0,ZJ 


























































6 






















(j 








































































































— 


i 















V- 2 = -f(ai + 3) 



9. 











— 


B J 




















































































^ 






















/ 













































r 


)(2,0)6 










/• 






l - 
































-F(-3,-3 

i 


■i 








/ 






i 
















— 


fi 















y + 3= §(x + 3) 



15. 



Alternate answer: y — |(x — 2) 













i 


/ 
































^ 


/ 


















































































/ 








/ 












/ 


s 


















& 


> 




y 


-A 


,c 


1 










(j 


-1 




























































y = ix 


















J 




' 




— 


fi 















»=!(* + 4) 



3.5. POINT-SLOPE FORM OF A LINE 



227 



17. 



v = f(a; + 3) 



21. 













6 •■ 






































































































































6 






V F 


>{ q rA 








(3 






7-r v u, u; 


























































2/ = 














J 








-If) J 















2 


/ 




7 




r 




7 




I 




i 


---fl(-4,- 


A ». 




=,k=; --. i 


-h ^ 


/-- 


f 


•--. = _ 


7 


"^-^ 


/ 6- 


' 



y+l = -|(x + 4) 



23. 



19. 



V= f(*+l) 













6 ( 






































































































































6 










- f 


f(- 


-1 


.( 


) 




6 














































































s 


s 












— 


fi 












\ 





\ 


\ 




— 


6 


< 
















\ 




















































































































6 






















(j 


































































\ 












pi ' q ' /i\ 


\ 














fiJ 


5 








\ 





25. 44.5 ft/s 



y + 4=f(x + 3) 



228 



CHAPTER 3. INTRODUCTION TO GRAPHING 



3.6 Standard Form of a Line 

In this section we will investigate the standard form of a line. Let's begin with 
a simple example. 



You Try It! 



Solve x — 2y = 6 for y and 
plot the result. 



EXAMPLE 1. Solve the equation 2x + 3y = 6 for y and plot the result. 

Solution: First we solve the equation 2x + 2>y = 6 for y. Begin by isolating 
all terms containing y on one side of the equation, moving or keeping all the 
remaining terms on the other side of the equation. 



2x + 3y = 6 
2x + 3y - 2x = 6 - 2x 
3y = 6 — 2x 
3y 6 — 2x 
T ~ 3 



Original equation. 

Subtract 2x from both sides. 

Simplify. 

Divide both sides by 3. 



Just as multiplication is dis- 
tributive with respect to ad- 
dition 

a{b + c) = ab + ac, 

so too is division distributive 
with respect to addition. 

a + b a b 
c c c 



When dividing a sum or a difference by a number, we use the distributive 
property and divide both terms by that number. 



On the left, simplify. On the right, 
divide both terms by 3. 



G 


2x 


3 


3 


9 - 


2x 




3 



Simplify. 



Finally, use the commutative property to switch the order of the terms on the 
right-hand side of the last result. 



Answer: 




2x 

y 



2 

-— x 

3 



Add the opposite. 

Use the commutative property to 
switch the order. 



Because the equation 2x + 3j/ = 6 is equivalent to the equation y = — -^x + 2, 
the graph of 2x + 3y = 6 is a line, having slope m = —2/3 and y-intercept 
(0, 2). Therefore, to draw the graph of 2x + 3y = 6, plot the y-intercept (0, 2), 
move down 2 and 3 to the right, then draw the line (see Figure 3.90). 

□ 



The form Ax + By = C, 
where either A = or B = 0, 
will be handled at the end of 
this section. 



In general, unless B = 0, we can always solve the equation Ax + By = C 



3.6. STANDARD FORM OF A LINE 



229 




*■ x 



Figure 3.90: The graph of 2x + 3y = 6, or equivalently, y = — ^x + 2. 



for y: 



Ax + By = 


c 


Ax + By - Ax = 


C- Ax 


By = 


C- Ax 


By 


C - Ax 


B 


B 


y = 


C Ax 
B ~ ~B 


y = 


A C 
~B X+ B 



Original equation. 

Subtract Ax from both sides. 

Simplify. 

Divide both sides by B, 
possible if B ^ 0. 

On the left, simplify. On the right 
distribute the B. 

Commutative property. 



Note that the last result is in slope-intercept form y = mx + b, whose graph is 
a line. We have established the following result. 



Fact. The graph of the equation Ax + By = C, is a line. 



Important points. A couple of important comments are in order. 

1. The form Ax + By = C requires that the coefficients A, B, and C are 
integers. So, for example, we would clear the fractions from the form 



1 2 1 

—x -\ — y = — 

2 3 y 4 



230 CHAPTER 3. INTRODUCTION TO GRAPHING 

by multiplying both sides by the least common denominator. 

i2 G* + I y ) = G) 12 

Qx + 8y = 3 
Note that the coefficients are now integers. 

2. The form Ax + By = C also requires that the first coefficient A is non- 
negative; i.e., A > 0. Thus, if we have 

— 5a; + 2y = 6, 

then we would multiply both sides by —1, arriving at: 

-l(-5x + 2y) = (6)(-l) 
5a; — 2y = —6 

Note that A = 5 is now greater than or equal to zero. 

3. If A, B, and C have a common divisor greater than 1, it is recommended 
that we divide both sides by the common divisor, thus "reducing" the 
coefficients. For example, if we have 

3x + 12y = -24, 

then dividing both side by 3 "reduces" the size of the coefficients. 

3a; + 12y _ -24 

3 ~ "ST 

x + 4y = -8 



Standard form. The form Ax + By = C, where A, B, and C are integers, 
and A > 0, is called the standard form of a line. 



Slope-Intercept to Standard Form 

We've already transformed a couple of equations in standard form into slope- 
intercept form. Let's reverse the process and place an equation in slope- 
intercept form into standard form. 



STANDARD FORM OF A LINE 



231 



You Try It! 



EXAMPLE 2. Given the graph of the line in Figure 3.91, find the equation Given the graph of the line 
of the line in standard form. below, find the equation of 

the line in standard form. 




Figure 3.91: Determine the equa- 
tion of the line. 




Figure 3.92: The line has y- 
intercept (0,-3) and slope —5/2. 




* x 



Solution: The line intercepts the j/-axis at (0,-3). From (0,-3), move 
up 5 units, then left 2 units. Thus, the line has slope Ay/Aa: = —5/2 (see 
Figure 3.92). Substitute —5/2 for m and —3 for b in the slope-intercept form 
of the line. 



y = rax + i 
5 



Slope-intercept form. 
Substitute: —5/2 for m, —3 for 



To put this result in standard form Ax + By = C, first clear the fractions by 
multiplying both sides by the common denominator. 



2y 

2y 

2y 



—x — 3 

2 



—x 

2 

-5a; — 6 



Multiply both sides by 2. 

2 [3] Distribute the 2. 

Multiply. 



That clears the fractions. To put this last result in the form Ax + By = C, we 
need to move the term —5x to the other side of the equation. 



5x + 2y 
5x + 2y 



-5a; — 6 + 5a; 
-6 



Add 5a; to both sides. 
Simplify. 



232 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Answer: 3a; 



Thus, the standard form of the line is 5x + 2y = —6. Note that all the coeffi- 
cients are integers and the terms are arranged in the order Ax + By = C, with 
A>0. 

□ 



You Try It! 



Find the standard form of 
the equation of the line that 
passes through the points 
(-2,4) and (3,-3). 



Point-Slope to Standard Form 

Let's do an example where we have to put the point-slope form of a line in 
standard form. 



EXAMPLE 3. Sketch the line passing through the points (—3,-4) and 
(1, 2), then find the equation of the line in standard form. 

Solution: Plot the points (—3, —4) and (1, 2), then draw a line through them 
(see Figure 3.93). 




(-3,-4j 



Figure 3.93: The line through (—3, —4) and (1,2). 



Use the points (—3,-4) and (1,2) to calculate the slope. 

Slope formula. 



Slope = — — 

Ax 



2 ~(-4) 
l-(-3) 

6 

4 
3 

2 



Subtract coordinates of (—3, —4) 
from the coordinates of (1, 2). 

Simplify. 
Reduce. 



STANDARD FORM OF A LINE 



233 



Let's substitute (xo,yo) = (1,2) and m = 3/2 in the point-slope form of the 
line. (Note: Substituting (xo,yo) = (—3,-4) and m = 3/2 would yield the 
same answer.) 



y-y = m (x - xq) 
1/-2 =|(a;-l) 



Point-slope form. 

Substitute: 3/2 for m, 1 for xo, 
and 2 for j/q. 



The question requests that our final answer be presented in standard form. 
First we clear the fractions. 



3 3 

y — 2 = —x 

y 2 2 



2 [y - 2] = 2 



2y - 2[2] = 2 



2y - 4 = 3a; - 3 



rs 


3" 






— X 

2 


~ 2 




"3 

— X 

2 


-2 




'3" 

2 



Distribute the 3/2. 
Multiply both sides by 2. 

Distribute the 2. 
Multiply. 



Now that we've cleared the fractions, we must order the terms in the form 
Ax + By = C . We need to move the term 3a; to the other side of the equation. 

1y — 4 — Sx = Sx — 3 — Sx Subtract 3a; from both sides. 

— Sx + 2y — 4 = —3 Simplify, changing the order on the 

left-hand side. 

To put this in the form Ax + By = C, we need to move the term —4 to the 
other side of the equation. 



-Sx + 2y-A + A = -3 + 4 
-3x + 2y = 1 



Add 4 to both sides. 
Simplify. 



It appears that —3a; + 2y = 1 is in the form Ax + By = C. However, standard 
form requires that A > 0. We have A = —3. To fix this, we multiply both 
sides by —1. 



If we fail to reduce the slope 
to lowest terms, then the 
equation of the line would be: 

y-2 = ^(x-l) 

Multiplying both sides by 4 
would give us the result 

Ay — 8 = 6a; — 6, 

or equivalently: 

-6a; + Ay = 2 

This doesn't look like the 
same answer, but if we divide 
both sides by —2, we do get 
the same result. 



3a; — 2y 



-1 



This shows the importance of 
requiring A > and "re- 
ducing" the coefficients A, B, 
and C. It allows us to com- 
pare our answer with our col- 
leagues or the answers pre- 
sented in this textbook. 



-1 [-3s + 2y] = -1 [1] 
3x- 2y = -1 



Multiply both sides by — 1. 
Distribute the — 1. 



Thus, the equation of the line in standard form is 3a; — 2y = — 1. 



Answer: 7x + 5y = 6 



□ 



234 



CHAPTER 3. INTRODUCTION TO GRAPHING 



Intercepts 

We've studied the y-intercept, the point where the graph crosses the y-axis, but 
equally important are the ^-intercepts, the points where the graph crosses the 
a;-axis. In Figure 3.94, the graph crosses the x-axis three times. Each of these 
crossing points is called an rc-intercept. Note that each of these x-intercepts 
has a y-coordinate equal to zero. This leads to the following rule. 



x-Intercepts. To find the x-intercepts of the graph of an equation, substitute 
y = into the equation and solve for x. 



Similarly, the graph in Figure 3.95 crosses the y-axis three times. Each of these 
crossing points is called a y-intercept. Note that each of these y-intercepts has 
an ^-coordinate equal to zero. This leads to the following rule. 

y-Intercepts. To find the y-intercepts of the graph of an equation, substitute 
x = into the equation and solve for y. 





Figure 3.94: Each x-intercept 
has a y-coordinate equal to 



Figure 3.95: Each y-intercept 
has an cc-coordinate equal to 
zero. 



Let's put these rules for finding intercepts to work. 



You Try It! 



Find the x- and y-intercepts 
of the line having equation 
3x + 4y = -12. Plot the 
intercepts and draw the line. 



EXAMPLE 4. Find the x- and y-intercepts of the line having equation 
2x — 3y = 6. Plot the intercepts and draw the line. 



STANDARD FORM OF A LINE 



235 



Solution: We know that the graph of 2x — 3y = 6 is a line. Furthermore, 
two points completely determine a line. This means that we need only plot the 
x- and y-intercepts, then draw a line through them. 



To find the ^-intercept of 2x — 3y 
6, substitute for y and solve for x. 



To find the y-intercept of 2x — 3y = 
6, substitute for x and solve for y. 



2x — 3y 
2a; -3(0) 

2x 
2x 
~2 
x 



6 
6 
6 
6 
2 
3 



2x - 


-3y = 


6 


2(0)- 


-3p = 


6 




-3y = 


6 




-3y 


6 




-3 


-3 




y = 


= -2 



Thus, the ^-intercept of the line is 
(3,0). 



Thus, the y-intercept of the line is 

(0,-2). 



Plot the x-intercept (3, 0) and the y-intercept (0, —2) and draw a line through 
them (see Figure 3.96). 




2x — 3y = 6 



* x 



Figure 3.96: The graph of 2x — 3y = 6 has intercepts (3, 0) and (0, —2) 



Answer: 



x-intercept: (—4, 0) 
y-intercept: (0, —3) 

V 











5 






















































( 


-4 


M 


11 












') 












-5- 


















5 






















fn 







N 












(0, ^ , 


>) 




















5 













3x + 4y = -12 



□ 



EXAMPLE 5. Sketch the line 4x + 3y = 12, then sketch the line through 
the point (—2, —2) that is perpendicular to the line 4x + 3y = 12. Find the 
equation of this perpendicular line. 

Solution: Let's first find the x- and y-intercepts of the line 4x + 3y = 12. 



You Try It! 



Find the equation of the line 
that passes through the 
point (3, 2) and is 
perpendicular to the line 
6x — by = 15. 



236 



CHAPTER 3. INTRODUCTION TO GRAPHING 



To find the x-intercept of the line 
Ax + 3y = 12, substitute for y and 
solve for x. 

Ax + 3y = 12 

4a; + 3(0) = 12 

Ax = 12 

Ax _ 12 

T = T 

x = 3 

Thus, the x-intercept of the line is 
(3,0). 



To find the y-intercept of the line 
Ax + 3y = 12, substitute for x and 
solve for y. 

Ax + 3y = 12 

4(0) + 3y = 12 

3j/ = 12 

3y = 12 

3 3 

y = A 

Thus, the j/-intercept of the line is 
(0,4). 



You could also solve for y to Plot the intercepts and draw a line through them. Note that it is clear from 
put 3a; + Ay = 12 in slope- the graph that the slope of the line 3a; + Ay = 12 is —4/3 (see Figure 3.97). 
intercept form in order to de- 
termine the slope. 

y y 




Figure 3.97: The graph of Ax+3y = 
12 has intercepts (3,0) and (0,4) 
and slope —4/3. 




* x 



Figure 3.98: The slope of the per- 
pendicular line is the negative re- 
ciprocal of —4/3, namely 3/4. 



Because the slope of 3x + Ay = 12 is —4/3, the slope of a line perpendicular 
to 3.t + Ay = 12 will be the negative reciprocal of —4/3, namely 3/4. Our 
perpendicular line has to pass through the point (—2, —2). Start at (—2, —2), 
move 3 units upward, then 4 units to the right, then draw the line. It should 
appear to be perpendicular to the line 3x + Ay = 12 (see Figure 3.98). 



Finally, use the point-slope form, m = 3/4, and (xo,yo) = (—2,-2) to 



STANDARD FORM OF A LINE 



237 



determine the equation of the perpendicular line. 



y-y Q = m(x - x ) 



Point-slope form. 



y - (-2) = -{x- (-2)) Substitute: 3/4 for to, -2 for x , 



y + 2 = -(x + 2) 



and —2 for j/o- 
Simplify. 



Let's place our answer in standard form. Clear the fractions. 

Distribute 3/4. 



3 6 
y + 2 = -x+- 



4 [y + 2] = 4 



r3 6i 




-x+ - 




[4 4j 




[3 1 




[fil 


T x 


+ 4 




[A 




[ij 



Multiply both sides by 4. 



Ay + 4 [2] = 4 -a; + 4 - Distribute the 4. 
4y + 8 = 3.T + 6 Multiply. 

Rearrange the terms to put them in the order Ax + By = C. 



Ay + 8 - 3a; = 3a; + 6 - 3x 

-3x + Ay + 8 = 6 
-3x + Ay + 8-8 = 6-8 
-3x + Ay = -2 
-l(-3x + Ay) = -l(-2) 
3x - Ay = 2 



Subtract 3a; from both sides. 

Simplify. Rearrange on the left. 

Subtract 8 from both sides. 

Simplify. 

Multiply both sides by —1. 

Distribute the — 1. 



Hence, the equation of the perpendicular line is 3a; — Ay = 2. 



Answer: 5x + 6y = 27 



□ 



Horizontal and Vertical Lines 

Here we keep an earlier promise to address what happens to the standard form 
Ax + By = C when either A = or B = 0. For example, the form 3a; = 6, 
when compared with the standard form Ax + By = C, has B = 0. Similarly, 
the form 2y = —12, when compared with the standard form Ax + By = C, 
has A = 0. Of course, 3a; = 6 can be simplified to x = 2 and 2y = — 12 can 
be simplified to y = —6. Thus, if either A = or B = 0, the standard form 
Ax + By = C takes the form x = a and y = b, respectively. 

As we will see in the next example, the form x = a produces a vertical line, 
while the form y = b produces a horizontal line. 



238 



CHAPTER 3. INTRODUCTION TO GRAPHING 



You Try It! 



Sketch the graphs of x 
and y = 2. 



EXAMPLE 6. Sketch the graphs of x = 3 and y = -3. 

Solution: To sketch the graph of x = 3, recall that the graph of an equation 
is the set of all points that satisfy the equation. Hence, to draw the graph of 
x = 3, we must plot all of the points that satisfy the equation x = 3; that is, 
we must plot all of the points that have an x-coordinate equal to 3. The result 
is shown in Figure 3.99. 

Secondly, to sketch the graph of y = —3, we plot all points having a y- 
coordinate equal to —3. The result is shown in Figure 3.100. 

y y 



->■ X 



*■ X 



Figure 3.99: The graph of x 
a vertical line. 



3 is 



Figure 3.100: The graph of y 
is a horizontal line. 



Things to note: A couple of comments are in order regarding the lines in 
Figures 3.99 and 3.100. 

1. The graph of x = 3 in Figure 3.99, being a vertical line, has undefined 
slope. Therefore, we cannot use either of the formulae y = mx + b or 
y — j/o = m(x — xq) to obtain the equation of the line. The only way we 
can obtain the equation is to note that the line is the set of all points 
(x, y) whose ^-coordinate equals 3. 

2. However, the graph of y = —3, being a horizontal line, has slope zero, 
so we can use the slope-intercept form to find the equation of the line. 
Note that the y-intercept of this graph is (0,-3). If we substitute these 
numbers into y = mx + b, we get: 



y = mx + b 
y = 0x + (-3) 
y = -3 



Slope-intercept form. 
Substitute: for m, —3 for b. 
Simplify. 



Answer: 



3.6. STANDARD FORM OF A LINE 



239 



However, it is far easier to just look at the line in Figures 3.100 and note 
that it is the collection of all points (x,y) with y = 3. 



x = -2 
t>-' 



y = 2 



-*■ x 



□ 



240 



CHAPTER 3. INTRODUCTION TO GRAPHING 



f> t* t* Exercises ■** -** ■** 



In Exercises 1-6, place the given standard form into slope-intercept form and sketch its graph. Label 
the graph with its slope-intercept form. 



1.4x-3y = 9 

2.2x-3y = 3 
3. 3x-2y = 6 



4. 5x - 3y = 3 

5. 2x + 3y = 12 

6. 3x + Ay = 8 



In Exercises 7-10, determine an equation of the given line in standard form. 
7. 9. 



8. 



y 













6- 


















































































































































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3.6. STANDARD FORM OF A LINE 



241 



In Exercises 11-16, sketch the line passing through the points P and Q. Label the line with its equation 
in standard form. 



11. P(-l,4) andQ(2,-4) 

12. P(-l,4) and (5(3,1) 

13. P(-l,-l) and Q(3,4) 



14. P(2,-l) andQ(4,4) 

15. P(-3,l) and Q(2, -3) 

16. P(-4,3) and Q(0,0) 



In Exercises 17-22, plot the x- and y-intercepts of the line having the given equation, then draw the 
line through the intercepts and label it with its equation. 



17. 2x - 5y = 10 

18. 2x + 3y = -6 

19. 3x-2y = 6 



20. 3x - Ay = 12 

21. 2x + 3y = 6 

22. 2x-3y = -6 



23. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 1,5) that is parallel to the line Ax + 
5y = -20. 

24. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 3,2) that is parallel to the line 3x + 
5j/ = -15. 

25. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 1,— 2) that is perpendicular to the 
line 5x + 2y = 10. 

26. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 1,— 2) that is perpendicular to the 
line 2x + 5y = -10. 



27. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 4, — 5) that is perpendicular to the 
line Ax + 3y = —12. 

28. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 2,— 3) that is perpendicular to the 
line 3x + 2y = 6. 

29. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 3,2) that is parallel to the line 5x + 
Ay = 20. 

30. Find an equation of the line (in stan- 
dard form) that passes through the point 
P(— 1,3) that is parallel to the line 3x + 
5y = -15. 



31 



32 



Sketch the equation of the horizontal line 
passing through the point P(— 5, — 4) and 
label it with its equation. 

Sketch the equation of the vertical line 
passing through the point P(— 4, 4) and 
label it with its equation. 



33 



34. 



Sketch the equation of the vertical line 
passing through the point P(— 2, —4) and 
label it with its equation. 

Sketch the equation of the vertical line 
passing through the point P(l,3) and la- 
bel it with its equation. 



242 



CHAPTER 3. INTRODUCTION TO GRAPHING 



£»• s*> £*■ Answers >•* >*$ •** 



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7. 4a; + 5y = 
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13. 











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4a; + 5y = -7 



3.6. STANDARD FORM OF A LINE 



243 



17. 



(0,-2)- 



2x - by = 10 



* x 



(5,0) 



23. 



Ax + 5y = 21 













tP(- 


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fi 


































































































































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Ax + 5y = -20 



19. 



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244 



CHAPTER 3. INTRODUCTION TO GRAPHING 



29. 



33. 



5x + 4y = -7 




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31. 



(-5,-4> 



-■y = —4 

JLJ 



Chapter 4 



Systems of Linear Equations 



In 1801, Carl Frederick Gauss (1777-1885) computed the orbit of the newly 
discovered planetoid Ceres from just a few observations by solving a system 
of equations. He invented a method (called Gaussian elimination) that is still 
used today. Solving systems of equations has been a subject of study in many 
other cultures. The ancient Chinese text Jiuzhang Suanshu (translated as 
Nine Chapters of Mathematical Art) written during the Han dynasty (206 
BC-220 AD) describes 246 problems related to practical situations such as 
land measurement, construction, and commerce. Here is one of the problems 
described in the text: "One pint of good wine costs 50 gold pieces, while one 
pint of poor wine costs 10. Two pints of wine are bought for 30 gold pieces. 
How much of each kind of wine was bought?" This problem can be solved 
by using a system of equations. In this chapter, we will learn how to solve 
systems of linear equations by using a variety of methods, including Gaussian 
elimination. 



245 



246 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 







8- 


































r — 3?/ = - 


y 












(3,4) 




























































































2x 


+ 


32/ 


= 


18 






















-9, 


















> 


< 


L_ 


2 





















4.1 Solving Systems by Graphing 

In this section we introduce a graphical technique for solving systems of two 
linear equations in two unknowns. As we saw in the previous chapter, if a point 
satisfies an equation, then that point lies on the graph of the equation. If we 
are looking for a point that satisfies two equations, then we are looking for a 
point that lies on the graphs of both equations; that is, we are looking for a 
point of intersection. 

For example, consider the the two equations 



Figure 4.1: The point of in- 
tersection is a solution of the 
system of linear equations. 



x 
2x 



32/ 
32/ 



-9 
18. 



which is called a system of linear equations. The equations are linear equations 
because their graphs are lines, as shown in Figure 4.1. Note that the two lines 
in Figure 4.1 intersect at the point (3,4). Therefore, the point (3,4) should 
satisfy both equations. Let's check. 

Substitute 3 for x and 4 for y. Substitute 3 for x and 4 for y. 



x - 3y = 


-9 


3 - 3(4) = 


-9 


3-12 = 


-9 


-9 = 


-9 



2a; + 32/ 

2(3) + 3(4) 
6+12 

18 



18 
18 
18 
18 



Hence, the point (3, 4) satisfies both equations and is called a solution of the 
system. 

Solution of a linear system. A point (x, y) is called a solution of a system 
of two linear equations if and only if it satisfied both equations. Furthermore, 
because a point satisfies an equation if and only if it lies on the graph of the 
equation, to solve a system of linear equations graphically, we need to determine 
the point of intersection of the two lines having the given equations. 



You Try It! 



Solve the following system of 
equations: 



2x 



by 
V 



-10 

x- 1 



Let's try an example. 



EXAMPLE 1. Solve the following system of equations: 

3x + 2y = 12 

y = x + 1 



(4.1) 



Solution: We are looking for the point (x, y) that satisfies both equations; 
that is, we are looking for the point that lies on the graph of both equations. 
Therefore, the logical approach is to plot the graphs of both lines, then identify 
the point of intersection. 

First, let's determine the x- and 2/-intercepts of 3a: + 2y = 12. 



4.1. SOLVING SYSTEMS BY GRAPHING 



247 



To find the ir-intercept, let y = 0. To find the y- intercept, let x = 0. 



3x + 2y 
3x + 2(0) : 
3x 



12 
12 
12 



3x + 2y = 12 

3(0) + 2y = 12 

2y=12 

y = 6 



Hence, the ^-intercept is (4,0) and the y-intercept is (0,6). These intercepts 
are plotted in Figure 4.2 and the line 3x + 2y = 12 is drawn through them. 

Comparing the second equation y = x + 1 with the slope-intercept form 
y = mx + b, we see that the slope is m = 1 and the y-intercept is (0, 1). Plot 
the intercept (0, 1), then go up 1 unit and right 1 unit, then draw the line (see 
Figure 4.3). 




3x + 2y = 12 

Figure 4.2: Drawing the graph of 
3x + 2y= 12. 




Figure 4.3: Drawing the graph of 

y = x + 1. 



We are trying to find the point that lies on both lines, so we plot both lines 
on the same coordinate system, labeling each with its equation (see Figure 4.4). 
It appears that the lines intersect at the point (2, 3), making (x, y) = (2, 3) the 
solution of System 4.1 (see Figure 4.4). 



Check: To show that (x 1 y) = (2, 3) is a solution of System 4.1, we must show 
that we get true statements when we substitute 2 for x and 3 for y in both 
equations of System 4.1. 



248 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Answer: (5,4) 




* x 



3x + 2y = 12 



Figure 4.4: The coordinates of the point of intersection is the solution of 
System 4.1. 



Substituting 2 for x and 3 for y in 
3x + 2y = 12, we get: 

3x + 2y = 12 

3(2) + 2(3) = 12 

6 + 6 = 12 

12 = 12 

Hence, (2, 3) satisfies the equation 
3x + 2y = 12. 



Substituting 2 for x and 3 for y in 
y = x + 1, we get: 

y = x + 1 
3 = 2 + 1 
3 = 3 



Hence, (2,3) satisfies the equation 
y = x + 1. 



Because (2, 3) satisfies both equations, this makes (2, 3) a solution of System 4.1. 



□ 



You Try It! 



Solve the following system of 
equations: 



-4x 



3y 
2y 



12 



EXAMPLE 2. Solve the following system of equations: 

3a; — by = — 15 

2x + y = -4 



(4.2) 



Solution: Once again, we are looking for the point that satisfies both equations 
of the System 4.2. Thus, we need to find the point that lies on the graphs of 
both lines represented by the equations of System 4.2. The approach will be to 
graph both lines, then approximate the coordinates of the point of intersection. 
First, let's determine the x- and y-intercepts of 3x — 5y = —15. 



4.1. SOLVING SYSTEMS BY GRAPHING 



249 



To find the x-intercept, let y = 0. To find the y-intercept, let x = 0. 



3x — by = 


-15 


32-5(0) = 


-15 


3x = 


-15 


x = 


-5 



3x - 


- by = 


-15 


3(0)- 


-by = 


-15 




-by = 


-15 




V = 


3 



Hence, the ^-intercept is (—5,0) and the y-intercept is (0,3). These intercepts 
are plotted in Figure 4.5 and the line 3a; — by = — 15 is drawn through them. 
Next, let's determine the intercepts of the second equation 2x + y = —4. 

To find the x-intercept, let y = 0. To find the y-intercept, let x = 0. 



2x + y = 


-4 


2a; + = 


-4 


2a; = 


-4 


x = 


-2 



2a; + y = 


-4 


2(0) +y = 


-4 


V = 


-4 



Hence, the a;-intercept is (—2, 0) and the y-intercept is (0, —4). These intercepts 
are plotted in Figure 4.6 and the line 2x -\- y = —4 is drawn through them. 



V 


3x — by - 








^ 




^ 








*7 


(0,3) . 


^ 




S 


^ 




.(-5,0) ^ . 




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J — 10- 





-15 



y 



y^j- r 10 J 




3 




v 




3 




v 




3 




v 




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v 




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-1U (— 2,UJ ^ 


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(0,-4) - 


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V 




3 




v 


J — 10^ 


3 



Later in this section we will 
learn how to use the intersect 
utility on the graphing 
calculator to obtain a much 
more accurate approximation 
of the actual solution. Then, 
in Sections 4.2 and 4.3, we'll 
show how to find the exact 
solution. 



Figure 4.5: 
the line 3a; - 



Drawing the graph of 
-by = -15. 



2a; + y = -4 

Figure 4.6: Drawing the graph of 
the line 2x + y = —4. 



To find the solution of System 4.2, we need to plot both lines on the same 
coordinate system and determine the coordinates of the point of intersection. 
Unlike Example 1, in this case we'll have to be content with an approximation 
of these coordinates. It appears that the coordinates of the point of intersection 
are approximately (—2.6, 1.4) (see Figure 4.7). 

Check: Because we only have an approximation of the solution of the system, 
we cannot expect the solution to check exactly in each equation. However, we 
do hope that the solution checks approximately. 



250 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



tkt r 1(p " 


■1:1; .)(/ - 


3 


: ^ 


v 


^ 


3 




v 


.^ 


3 


__,' 


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(-2.6,1.4): r^ : 




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2x 



y 



Figure 4.7: The approximate coordinates of the point of intersection are 
(-2.6,1.4). 



Substitute (x, y) = (—2.6,1.4) into 
the first equation of System 4.2. 



3a; — 5y = 


-15 


3(-2.6) - 5(1.4) = 


-15 


-7.8-7 = 


-15 


-14.8 = 


-15 



Substitute {x,y) = (—2.6,1.4) into 
the second equation of System 4.2. 



2x + y = -4 




2(-2.6) + 1.4 = 


-4 


-5.2+1.4 = 


-4 


-3.8 = 


-4 



Approximate answer: 
(-2.7,-0.4) 



Note that (x,y) = (-2.6,1.4) does 
not check exactly, but it is pretty 
close to being a true statement. 



Again, note that (a;, y) = 
(—2.6,1.4) does not check ex- 
actly, but it is pretty close to being 
a true statement. 



Because (x,y) = (—2.6,1.4) very nearly makes both equations a true state- 
ment, it seems that (x, y) = (—2.6, 1.4) is a reasonable approximation for the 
solution of System 4.2. 

□ 



You Try It! 



Exceptional Cases 

Most of the time, given the graphs of two lines, they will intersect in exactly 
one point. But there are two exceptions to this general scenario. 



Solve the following system of EXAMPLE 3. Solve the following system of equations: 
equations: 



x-y 
-2x + 2y 



2x + 3y 
2x + 3y 



6 



-6 



(4.3) 



4.1. SOLVING SYSTEMS BY GRAPHING 



251 



Solution: Let's place each equation in slope-intercept form by solving each 
equation for y. 



Solve 2x + 3y = 6 for y: 

2x + 3y = 6 
2x + 3y-2x = 6-2x 
3y = 6 — 2x 
3y 6 — 2x 

~3 ~ 3 

2 

y = -o x + 



Solve 2x + 3y = —6 for y: 



2x + 3y = 


-6 


2x + 3y — 2x = 


-6 -2a; 


3y = 


-6- 2x 


3y 


-6 -2x 


3 


3 



y 



Comparing y = (— 2/3)a; + 2 with the slope-intercept form y = mx + b tells 
us that the slope is m = —2/3 and the y-intercept is (0, 2). Plot the intercept 
(0, 2), then go down 2 units and right 3 units and draw the line (see Figure 4.8). 

Comparing y = (— 2/3)a: — 2 with the slope-intercept form y = mx + b 
tells us that the slope is m = —2/3 and the y-intercept is (0,-2). Plot the 
intercept (0, —2), then go down 2 units and right 3 units and draw the line (see 
Figure 4.9). 



2x + 3y 





Figure 4.8: Drawing the graph of 
the line 2x + 3y = 6. 



Figure 4.9: Drawing the graph of 
the line 2x + 3y = —6. 



To find the solution of System 4.3, draw both lines on the same coordinate 
system (see Figure 4.10). Note how the lines appear to be parallel (they don't 
intersect). The fact that both lines have the same slope —2/3 confirms our 
suspicion that the lines are parallel. However, note that the lines have differ- 
ent jy-intercepts. Hence, we are looking at two parallel but distinct lines (see 
Figure 4.10) that do not intersect. Hence, System 4.3 has no solution. 



252 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



2x + 3y = 6 




*■ x 



2x + 3y = -6 
Figure 4.10: The lines 2x + 3y = 6 and 2x + 3y = —6 are parallel, distinct lines. 



Answer: No solution. 



□ 



You Try It! 



Solve the following system of EXAMPLE 4. Solve the following system of equations: 
equations: 

x-y = 3 

-2x + 2y = -6 



-6x + 3y = -12 

2x - y = 4 



(4.4) 




Figure 4.11: x — y = 3 and 
— 2x + 2 y = — 6 are the same 
line. 



Solution: Let's solve both equations for y. 



Solve x — y = 3 for y: 





x-y = 3 


X 


- y — x = 3 — x 




—y = —x + 3 




-l(-y) = -l(-x + 3) 



Solve 


-2x + 2y = - 


6 for y: 




-2x + 2y 


= -6 




-2x + 2y + 2x 


= -6 + 2.x 




2y 


= 2a; -6 




2y 


2x -6 




2 


2 




V 


= x — 3 



y = x 



Both lines have slope m = 1, and both have the same y-intercept (0,-3). 
Hence, the two lines are identical (see Figure 4.11). Hence, System 4.4 has an 
infinite number of points of intersection. Any point on either line is a solution 
of the system. Examples of points of intersection (solutions satisfying both 
equations) are (0,-3), (1,-2), and (3,0). 



4.1. SOLVING SYSTEMS BY GRAPHING 



253 



Alternate solution: A much easier approach is to note that if we divide 
both sides of the second equation —2x + 2y = —6 by —2, we get: 



Second equation in System 4.4. 
Divide both sides by —2. 

Distribute —2. 
Simplify. 



-2x + 2y = 


-6 


-2x + 2y 


-6 


-2 


-2 


-2x 2y 
-2 + -2 ~ 


-6 

^2 



x-y 



Hence, the second equation in System 4.4 is identical to the first. Thus, there 
are an infinite number of solutions. Any point on either line is a solution. 



Answer: There are an 
infinite number of solutions. 
The lines are identical, so 
any point on either line is a 
solution. 



Examples 1, 2, 3, and 4 lead us to the following conclusion. 

Number of solutions of a linear system. When dealing with a system of 
two linear equations in two unknowns, there are only three possibilities: 

1. There is exactly one solution. 

2. There are no solutions. 

3. There are an infinite number of solutions. 



□ 



Solving Systems with the Graphing Calculator 

We've already had experience graphing equations with the graphing calculator. 
We've also used the TRACE button to estimate points of intersection. How- 
ever, the graphing calculator has a much more sophisticated tool for finding 
points of intersection. In the next example we'll use the graphing calculator to 
find the solution of System 4.1 of Example 1. 



You Try It! 



EXAMPLE 5. Use the graphing calculator to solve the following system of Solve the following system of 

equations: equations: 

3x + 2?/=12 (45) 2x-5y = 9 

V = x + 1 y = 2x-5 



Solution: To enter an equation in the Y= menu, the equation must first be 



254 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



solved for y. Hence, we must first solve 3a; + 2y = 12 for y. 



Having the calculator ask 
"First curve," "Second 
curve," when there are only 
two curves on the screen may 
seem annoying. However, 
imagine the situation when 
there are three or more 
curves on the screen. Then 
these questions make good 
sense. You can change your 
selection of "First curve" or 
"Second curve" by using the 
up-and-down arrow keys to 
move the cursor to a 
different curve. 



3a; 



2y = 


12 




2y = 


12- 


- 3a; 


2y 


12 


— 3a; 


2 




2 




12 


3.r 


y = 


T 


~ T 



y 



Original equation. 

Subtract 3a; from both sides of the equation. 

Divide both sides by 2. 

On the left, simplify. On the right, 
distribute division by 2. 

Simplify. 



We can now substitute both equations of System 4.5 into the Y= menu (see 
Figure 4.12). Select 6:ZStandard from the ZOOM menu to produce the 
graphs shown in Figure 4.13. 



Ploti Plots 


Plot3 


sViB6-3^2*X 


■^eBX+1 




\Vi=W 




\Yh = 




■^Ye = 




^Vfi = 




sY? = 






Figure 4.12: Enter System 4.5 
equations into the Y= menu. 



Figure 4.13: Select 6:ZStandard 
to produce the graphs of the sys- 
tem (4.5) equations. 



The question now becomes "How do we calculate the coordinates of the 
point of intersection?" Look on your calculator case just above the TRACE 
button on the top row of buttons, where you'll see the word CA1C, painted in 
the same color as the 2ND key. Press the 2ND key, then the TRACE button, 
which will open the CALCULATE menu shown in Figure 4.14. 

Select 5:intersect. The result is shown in Figure 4.15. The calculator has 
placed the cursor on the curve y = 6 — (3/2)x (see upper left corner of your 
viewing screen), and in the lower left corner the calculator is asking you if you 
want to use the selected curve as the "First curve." Answer "yes" by pressing 
the ENTER button. 

The calculator responds as shown Figure 4.16. The cursor jumps to the 
curve y = x + 1 (see upper left corner of your viewing window), and in the 
lower left corner the calculator is asking you if you want to use the selected 
curve as the "Second curve." Answer "yes" by pressing the ENTER key again. 

The calculator responds as shown Figure 4.17, asking you to "Guess." In 
this case, leave the cursor where it is and press the ENTER key again to signal 
the calculator that you are making a guess at the current position of the cursor. 

The result of pressing ENTER to the "Guess" question in Figure 4.17 is 
shown in Figure 4.18, where the calculator now provides an approximation 



4.1. SOLVING SYSTEMS BY GRAPHING 



255 



MMMslla 



ualue 
2Tzero 
3:nininun 
4:naxinur v i 
5: intersect 

7:XfCx>dx 



Figure 4.14: Press 2ND, then 
TRACE to open the CALCU- 
LATE menu. Then select 5:in- 
tersect to produce the screen in 
Figure 4.15. 



V1=G-3/£*KF y^ 

i i y^ 


FiKSt CUKV4? 
H=0 


v=6 -■ 



Figure 4.15: Press the ENTER key 
on your calculator to say "yes" to 
the "First curve" selection. 



VE=H+i 




Second curve? 

H=0 IV: 



Figure 4.16: Press the ENTER key 
on your calculator to say "yes" to 
the "Second curve" selection. 



VE=H+i V 




Guess? 
H=0 


v=i -■ 



Figure 4.17: Press the ENTER key 
to signal the calculator that you are 
satisfied with the current position 
of the cursor as your guess. 



of the the coordinates of the intersection point on the bottom edge of the 
viewing window. Note that the calculator has placed the cursor on the point 
of intersection in Figure 4.17 and reports that the approximate coordinates of 
the point of intersection are (2, 3). 







Intersection 
K=2 


V=3 -■ 



Figure 4.18: Read the approximate coordinates of the point of intersection 
along the bottom edge of the viewing window. 



In later sections, when we 
investigate the intersection of 
two graphs having more than 
one point of intersection, 
guessing will become more 
important. In those future 
cases, we'll need to use the 
left-and-right arrow keys to 
move the cursor near the 
point of intersection we wish 
the calculator to find. 



256 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Reporting your solution on your homework. In reporting your solution 
on your homework paper, follow the Calculator Submission Guidelines from 
Chapter 3, Section 2. Make an accurate copy of the image shown in your 
viewing window. Label your axes x and y. At the end of each axis, put 
the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your 
calculator's WINDOW menu. Use a ruler to draw the lines and label each with 
their equations. Finally, label the point of intersection with its coordinates 
(see Figure 4.19). Unless instructed otherwise, always report every single digit 
displayed on your calculator. 




-10 + 



3x + 2y = 12 
Figure 4.19: Reporting your result on your homework paper. 



Approximate answer: 

(2,-1) 



□ 



You Try It! 



Solve the following system of 
equations: 



y 



6 



Sometimes you will need to adjust the parameters in the WINDOW menu 
so that the point of intersection is visible in the viewing window. 



EXAMPLE 6. Use the graphing calculator to find an approximate solution 
of the following system: 

y = -~x + 7 

3 7 ^ 6 ) 

y = —x — 5 
y 5 

Solution: Each equation of System 4.6 is already solved for y, so we can 
proceed directly and enter them in the Y= menu, as shown in Figure 4.20. 
Select 6:ZStandard from the ZOOM menu to produce the image shown in 
Figure 4.21. 



4.1. SOLVING SYSTEMS BY GRAPHING 



257 



Ploti Plots Plots 
WiB-2/7*X+7 
W2B3/-5+X-5 
^Ys = 
\Vh = 

^7 = 




Figure 4.20: Enter the equations of 
System 4.6. 



Figure 4.21: Select 6:ZStandard 
to produce this window. 



Obviously, the point of intersection is off the screen to the right, so we'll have 
to increase the value of Xmax (set Xmax = 20) as shown in Figure 4.22. Once 
you have made that change to Xmax, press the GRAPH button to produce 
the image shown in Figure 4.23. 



WINDOW 
Xnun=-10 
Xnax=2@ 
Xscl=l 
Wiin=-10 
Wiax=l@ 
Vscl=l 

■i-Xres=l 




Figure 4.22: Change xmax to 20. 



Figure 4.23: Press the GRAPH 
button to produce this window. 



Now that the point of intersection is visible in the viewing window, press 
2ND CALC and select 5:intersect from the CALCULATE menu (see Figure 4.25) 
Make three consecutive presses of the ENTER button to respond to "First 
curve," "Second curve," and "Guess." The calculator responds with the im- 
age in Figure 4.25. Thus, the solution of System 4.6 is approximately (x, y) w 
(13.54837,3.1290323). 

Warning! Your calculator is an approximating machine. It is quite likely that 
your solutions might differ slightly from the solution presented in Figure 4.25 
in the last 2-3 places. 



Reporting your solution on your homework. In reporting your solution 
on your homework paper, follow the Calculator Submission Guidelines from 
Chapter 3, Section 2. Make an accurate copy of the image shown in your 
viewing window. Label your axes x and y. At the end of each axis, put 
the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your 



258 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



MMMslla 



ualue 
2Tzero 
3:nininun 
4:naxinur v i 
5: intersect 

7:XfCx>dx 



Figure 4.24: Press 2ND CALC 
to open the CALCULATE menu. 
Select 5:intersect to find the point 
of intersection. 




Intersection 
K=13.£HB3B7 



3.i£903£3 



Figure 4.25: Three consecutive 
presses of the ENTER key produce 
the coordinates shown at the bot- 
tom of the viewing window. 



calculator's WINDOW menu. Use a ruler to draw the lines and label each with 
their equations. Finally, label the point of intersection with its coordinates 
(see Figure 4.26). Unless instructed otherwise, always report every single digit 
displayed on your calculator. 




(13.548387,3.1290323) 



Figure 4.26: Reporting your result on your homework paper. 



Approximate answer: 
(-4.2,-0.4) 



□ 



4.1. SOLVING SYSTEMS BY GRAPHING 



259 



i*. ;». ;». 



Exercises 



■*j ■*-: •*; 



In Exercises 1-6, solve each of the given systems by sketching the lines represented by each equation 
in the system, then determining the coordinates of the point of intersection. Each of these problems 
have been designed so that the coordinates of the intersection point are integers. Check your solution. 



3a; 


-Ay-- 


= 24 




y- 


1 

= x - 

2 


x — 


Ay = 


-8 




y = 


5 ,« 
-x + 

4 


2x + y = 


6 




y = 


x + 3 



x- 2y 



y = 2 X + 6 



x + 2y = —6 

y = —3a; — . 

x — 3y = 6 

y = 2x - 7 



In Exercises 7-18, solve each of the given systems by sketching the lines represented by each equation 
of the given system on graph paper, then estimating the coordinates of the point of intersection to the 
nearest tenth. Check the solution. 



10. 



11. 



12. 



—x - 


-3y = 


-- 3 


x - 


-Ay = 


---A 


Ax - 


-3y = 


= -12 


—x - 


-4y = 


= 4 


— 3x + 3y 


= -9 


— 3x + 3y 


= -12 


X 


- y = 


-2 


2x- 


-2y = 


6 


Qx - 


-7y = 


-42 




y = 


1 

-A X + 


Ax + 3y = 


24 




y = 


1 

-a; + 5 



13. 



14. 



15. 



16. 



17. 



18. 



6x-7y= -42 

y 

7x — 8y = 56 

y = x — A 



1 



6x + 3y = 12 
-2a; - y = A 

x - Ay = -A 
-x + Ay = -A 

3x + y = 3 
-2x + 3y = -6 

2x-y = -2 
-x - 2y = A 



260 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



In Exercises 19-24, use the graphing calculator to solve the given system. Round your answer to the 
nearest tenth. Use the Calculator Submission Guidelines from Chapter 3, Section 2, when reporting 
your answer on your homework. 



19. 



20. 



21. 



y = 


-x + 7 
4 


y = 


-\x + 2 


y = 


-x + 6 
6 



x + 3 



y = —x — 3 
J 3 

4 
v = —-X— 1 



22. 



23. 



24. 



y = —x — 3 
y 3 



y 

y 
y 

y 

y 



--x-2 

8 

L 

-x + 1 

) 

--x + 5 

-a; + 3 
5 



-x — 5 



6 



In Exercises 25-30, use the graphing calculator to solve the given system. Round your answer to 
the nearest tenth. Use the Calculator Submission Guidelines when reporting your answer on your 
homework. 



25. 6x + 16y = 96 

-6x + 13y = -78 

26. -Ax + 16y = -64 

5x + 8y = 40 

27. -2z-llj/ = 22 

8a: - 12y = -96 



28. 



29. 



30. 



6a; — lOy = 


60 


2a; - 18 j/ = 


-36 


-6a; + 2y = 


-12 


12a; + 3y = 


-36 


— 3a; + y = 


-3 


14a; + 3y = 


-42 



j*- j*- ?*- Answers •** ~*^ •** 



1.(4,-3) 

3. (1,4) 

5. (-2,-2) 

7. (-3.4,0.1) 

9. No solution. Lines are parallel. 



11. (-1.8,4.5) 

13. (-3.8,2.8) 

15. No solution. Lines are parallel. 

17. (1.4,-1.1) 

19. (-4.6,3.5) 



4.1. SOLVING SYSTEMS BY GRAPHING 261 

21.(1.1,-1.6) 27. (-11.8,0.1) 

23.(6.7,2.1) 29.(6.0,12.0) 

25. (14.3,0.6) 



262 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



4.2 Solving Systems by Substitution 

In this section we introduce an algebraic technique for solving systems of two 
equations in two unknowns called the substitution method. The substitution 
method is fairly straightforward to use. First, you solve either equation for 
either variable, then substitute the result into the other equation. The result 
is an equation in a single variable. Solve that equation, then substitute the 
result into any of the other equations to find the remaining unknown variable. 



You Try It! 



Solve the following system of EXAMPLE 1. Solve the following system of equations: 
equations: 

2x-5y = -8 

= 3x-l 



9x + 2y 
V 



-19 

13 + 3x 



V 



(4.7) 
(4.8) 



3x- 1 













5 






2x — by 
































































(1,2 






























R 


















c 














/ 
























/ 
























/ 






















/ 


R 















Figure 4.27: 2x - 5y = -8 
and y = x — 3 intersect at 
(1,2). 



Solution: Equation (4.8) is already solved for y. Substitute equation (4.8) 
into equation (4.7). This means we will substitute 3a: — 1 for y in equation (4.7). 

Equation (4.7). 

Substitute 3x — 1 for y in (4.7). 



Distribute —5. 

Simplify. 

Subtract 5 from both sides. 

Divide both sides by —13. 



As we saw in Solving Systems by Graphing, the solution to the system is the 
point of intersection of the two lines represented by the equations in the system. 
This means that we can substitute the answer x = 1 into either equation to find 
the corresponding value of y. We choose to substitute 1 for x in equation (4.8), 
then solve for y 7 but you will get exactly the same result if you substitute 1 for 
x in equation (4.7). 





2x — 5y = 


-8 


2x- 


5(3x - 1) = 


-8 


Now solve for 


X. 




2x - 


- 15a; + 5 = 


-8 




-13a; + 5 = 


-8 




-13a; = 


-13 




x = 


1 



3a; — 

3(1) 
2 



Equation (4.8). 
Substitute 1 for x. 
Simplify. 



Hence, (x,y) = (1, 2) is the solution of the system. 

Check: To show that the solution (x,y) = (1,2) is a solution of the system, 
we need to show that (x,y) = (1,2) satisfies both equations (4.7) and (4.8). 



4.2. SOLVING SYSTEMS BY SUBSTITUTION 



263 



Substitute (x,y) = (1,2) in equa- 
tion (4.7): 

2x - by = -8 

2(1) -5(2) = -8 

2- 10 = -8 



Thus, (1,2) satisfies equation (4.7). 



Substitute (x,y) = (1,2) in equa- 
tion (4.8): 

y = 3x — 1 
2 = 3(1)- 1 
2 = 3-1 
2 = 2 

Thus, (1,2) satisfies equation (4.8). 



Because (x,y) = (1,2) satisfies both equations, it is a solution of the system. Answer: (—3,4) 



Substitution method. The substitution method involves these steps: 

1. Solve either equation for either variable. 

2. Substitute the result from step one into the other equation. Solve the 
resulting equation. 

3. Substitute the result from step two into either of the original system equa- 
tions or the resulting equation from step one (whichever seems easiest), 
then solve to find the remaining unknown variable. 



□ 



EXAMPLE 2. Solve the following system of equations: 

bx-2y = 12 
Ax + y = 6 



(4.9) 
(4.10) 





You Try It! 








Solve the following system of 
equations: 




x- 2y = 13 
Ax - 3y = 26 





Solution: The first step is to solve either equation for either variable. This 
means that we can solve the first equation for x or y, but it also means that we 
could first solve the second equation for x or y. Of these four possible choices, 
solving the second equation (4.10) for y seems the easiest way to start. 



4a; + y = 6 

y = 6 — Ax 



Equation (4.10). 

Subtract Ax from both sides. 



264 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



5x-2y = 12 











5 J 


\ 








/ 












\ 






1 














\ 






/ 
















\ 




/ 
















\ 


1 






- 5 ( 


21/ 










\ 


1 




5 


13, -18/12 


)$ 




















/ 


\ 
















/ 


1 


\ 














5- 


/ 




\ 







Ax + y = 6 

Figure 4.28: 5x - 2y = 12 
and Ax + y = 6 intersect at 
(24/13,-18/13). 



Next, substitute 6 — Ax for y in equation (4.9). 



5x-2y= 12 

5x - 2(6 - Ax) = 12 

bx - 12 + 8x = 12 

13a;- 12 = 12 

13a; = 24 

_ 24 

X ~ 13 



Equation (4.9). 

Substitute 6 — 4a; for y in (4.9). 

Distribute —2. 

Simplify. 

Add 12 to both sides. 

Divide both sides by 13. 



Finally, to find the y-value, substitute 24/13 for x in the equation y = 6 — Ax 
(you can also substitute 24/13 for x in equations (4.9) or (4.10)). 



y = 


6- Ax 


y = 


6-4 


/24 
V13 


y = 


78 
13 ~ 


96 
13 


y = 


18 
~13 





Substitute 24/13 for x in y = 6 — Ax. 

Multiply, then make equivalent fractions. 
Simplify. 

Hence, (x, y) = (24/13, —18/13) is the solution of the system. 

Check: Let's use the graphing calculator to check the solution. First, we store 
24/13 in X with the following keystrokes (see the result in Figure 4.29). 



00BQ0 



Next, we store —18/13 in the variable Y with the following keystrokes (see the 
result in Figure 4.29). 



EDQHHQS 



ALPHA 



□ 



ENTER 



Now, clear the calculator screen by pressing the CLEAR button, then enter 
the left-hand side of equation 4.9 with the following keystrokes (see the result 
in Figure 4.30). 



m 



X, T, 9, n 



m 



ALPHA 



□ 



ENTER 



Now enter the left-hand side of equation 4.10 with the following keystrokes 
(see the result in Figure 4.30). Note that each left-hand side produces the 
number on the right-hand sides of equations (4.9) and (4.10). Thus, the solution 
(x, y) = (24/13,-18/13) checks. 



4.2. SOLVING SYSTEMS BY SUBSTITUTION 



265 



□ 



ALPHA 



□ 



ENTER 



1.846153846 
-18^13+V 

-1.384615385 



Figure 4.29: Storing 24/13 and 
-18/13 in X and Y. 



5*X-2*V 
4+X+V 



12 
6 



Figure 4.30: Checking equa- 
tions (4.9) and (4.10). 



Answer: (13/5,-26/5) 



□ 



EXAMPLE 3. Solve the following system of equations: 

3x — 2y = 6 
Ax + 5y = 20 







You Try It! 










(4.11) 
(4.12) 


Solve the following system of 
equations: 

3a; — 5y = 3 
5x — 6y = 2 



Solution: Dividing by —2 gives easier fractions to deal with than dividing by 
3, 4, or 5, so let's start by solving equation (4.11) for y. 



3x - 


-2y = 6 




— 2y = 6 — 3a; 




6 — 3a; 

y -2 




y = -3 + -x 



Equation (4.11). 

Subtract 3x from both sides. 

Divide both sides by —2. 

Divide both 6 and —3a; by —2 
using distributive property. 



266 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Substitute —3 + |x for y in equation (4.12). 



4x + 5y = 20 











f> 
























































(70/23,36/23)") 


























-5 


















5 










/ 




















/ 


















/ 


















/ 


5 













3x - 2y = 6 

Figure 4.31: 3x — 2y = 6 
and 4x + 5y = 20 intersect at 
(70/23,36/23). 



4x + 5 



4x + 5y 
„ 3 



—a; 
2 



20 
20 



Equation (4.12). 



Substitute 



-3 H — x for y. 
2 



4x- 15- 
8x - 30 



15 

— x = 20 
2 



15x 
23x 



40 

70 
70 
23 



Distribute the 5. 

Clear fractions by multiplying 
both sides by 2. 

Simplify. Add 30 to both sides. 
Divide both sides by 23. 



To find y, substitute 70/23 for x into equation y 



You could also 



substitute 70/23 for x in equations (4.11) or (4.12) and get the same result. 



y = 


-3 + 3 -x 


v = 


3/70 
" 3 + U23 


y = 


69 105 
~23 + _ 23" 


y = 


36 

23 



Substitute 70/23 for x. 

Multiply. Make equivalent fractions. 

Simplify. 

Hence, (x, y) = (70/23, 36/23) is the solution of the system. 

Check: To check this solution, let's use the graphing calculator to find the 
solution of the system. We already know that 3x — 2y = 6 is equivalent to 
y = —3 + |x. Let's also solve equation (4.12) for y. 



4x 



5y 
by 

y 
y 



20 

20 -Ax 

20 -4.x 



, 4 

4 x 

5 



Equation (4.12). 

Subtract 4x from both sides. 

Divide both sides by 5. 

Divide both 20 and — 4x by 5 
using the distributive property. 



Enter y = — 3+ |x and y = 4— $x into the Y= menu of the graphing calculator 
(see Figure 4.32). Press the ZOOM button and select 6:ZStandard. Press 
2ND CALC to open the CALCULATE menu, select 5:intersect, then press 
the ENTER key three times in succession to enter "Yes" to the queries "First 
curve," "Second curve," and "Guess." The result is shown in Figure 4.33. 
At the bottom of the viewing window in Figure 4.33, note how the coordinates 
of the point of intersection are stored in the variables X and Y. Without 



4.2. SOLVING SYSTEMS BY SUBSTITUTION 



267 



Ploti Plots Plots 
WiB-3+3/2*X 
^eB4-4^5*X 

^7 = 



Figure 4.32: Enter y 



y = 4 
tively. 



3 + | a; and 
\x in Yl and Y2, respec- 







Intersection 
H=3.0H3H?B3 


Y=i.Efi££lFH 



Figure 4.33: Use 5:intersect on 
the CALC menu to calculate the 
point of intersection. 



moving the cursor, (the variables X and Y contain the coordinates of the 
cursor), quit the viewing window by pressing 2ND QUIT, which is located 
above the MODE key. Then press the CLEAR button to clear the calculator 
screen. 

Now press the X,T,#,n key, then the MATH button on your calculator: 



This will open the MATH menu on your calculator (see Figure 4.34). Select 
l:^Frac, then press the ENTER key to produce the fractional equivalent of 
the decimal content of the variable X (see Figure 4.35). 



Hi 


U£ MUM 


CPX 


PRB 


Qe 


[►Frac 






2 = 


► Dec 






3: 


3 






4: 


3-K 






5: 


*f 






6'. 


fMinC 






7-i-fMaxC 







Figure 4.34: The MATH menu. 



X^Frac 
V^Frac 



70/23 
36/23 



Figure 4.35: Changing the contents 
of the variables X and Y to frac- 
tions. 



Repeat the procedure for the variable Y. Enter: 



ALPHA 



□ 



Select 1 : ►Frac, then press the ENTER key to produce the fractional equivalent 
of the decimal content of the variable Y (see Figure 4.35). Note that the 
fractional equivalents for X and Y are 70/23 and 36/23, precisely the same 
answers we got with the substitution method above. 



Answer: (-8/7,-9/7) 



□ 



268 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



You Try It! 



Solve the following system of 
equations: 



x = —v 



7 



6x — 8y 




Figure 4.36: 2x + 3y = 6 and 



solution. 



4 are parallel. No 



Answer: no solution 



Exceptional Cases Revisited 

It is entirely possible that you might apply the substitution method to a system 
of equations that either have an infinite number of solutions or no solutions at 
all. Let's see what happens should you do that. 



EXAMPLE 4. Solve the following system of equations: 

2x + 3y = 6 

2 

y = — —x + 4 



2 

-— x 

3 



(4.13) 
(4.14) 



Solution: Equation (4.14) is already solved for y, so let's substitute 
for y in equation (4.13). 



2x 



2x + 3y 


= 6 


Equation (4.13). 


3 {-r +i ) 


= 6 


2 
Substitute — x 


2x- 2x+ 12 


= 6 


Distribute the 3. 


12 


= 6 


Simplify. 



4 for y. 



Goodness! What happened to the x? How are we supposed to solve for x in 
this situation? However, note that the resulting statement, 12 = 6, is false, no 
matter what we use for x and y. This should give us a clue that there are no 
solutions. Perhaps we are dealing with parallel lines? 

Let's solve equation (4.13) for y, putting the equation into slope-intercept 
form, to help determine the situation. 



2a;- 



3y 
3y: 

y 



6 



-2x- 

2 
--x 



Equation (4.13). 

Subtract 2x from both sides. 

Divide both sides by 3. 



Thus, our system is equivalent to the following two equations. 

2 



y = 


--x + 2 


y = 


2 
— -x + 4 



(4.15) 
(4.16) 



These lines have the same slope —2/3, but different y-intercepts (one has y- 
intercept (0,2), the other has y-intercept (0,4)). Hence, these are two distinct 
parallel lines and the system has no solution. 



□ 



4.2. SOLVING SYSTEMS BY SUBSTITUTION 



269 



EXAMPLE 5. Solve the following system of equations: 

2x-6y = -8 

x = 3y — 4 



Solution: Equation (4.18) is already solved for x, so let's substitute 3y — 4 for 
x in equation (4.17). 







You Try It! 












Solve the following system of 


(4.17) 


equations: 


(4.18) 


-2Sx+Uy = -126 


A fnr 




y = 2x- 


9 



2x-6y = -8 
2(3y - 4) - 6y = -8 

6y-8-6y = -8 

-8 = -8 



Equation (4.17). 
Substitute 3y — 4 for x. 
Distribute the 2. 
Simplify. 



Goodness! What happened to the xl How are we supposed to solve for x in 
this situation? However, note that the resulting statement, —8 = —8, is a true 
statement this time. Perhaps this is an indication that we are dealing with the 
same line? 

Let's put both equations (4.17) and (4.18) into slope-intercept form so that 
we can compare them. 



Solve equation (4.17) for y: 



2x 



Solve equation (4.18) for y: 



6y = 


-8 




6y = 


-2x- 


-8 




-2a;- 


-8 


V 


-e 




y = 


i 


4 
3 



x - 


= 3y 


-4 


x + 4 = 


= 3y 




x + 4 
3 


= y 




y = 


l 

= —X 

3 


4 
+ 3 



Hence, the lines have the same slope and the same y-intercept and they are 
exactly the same lines. Thus, there are an infinite number of solutions. Indeed, 
any point on either line is a solution. Examples of solution points are (—4,0), 
(-1,1), and (2,2). 













y 
























L 














,i 




■Vj 




1 






(- 


1, 


















L) 








(2 


,2 




("4, 


') 


















5 


















5 




_2x- 


6« = 


- 


A 






















































s 













Figure 4.37: 2x - 6y = -8 
and x = 3y — 4 are the same 
line. Infinite number of solu- 
tions. 



Answer: There are an 
infinite number of solutions. 
Examples of solution points 
are (0,-9), (5,1), and 
(-3,-15). 



□ 



Tip. When you substitute one equation into another and the variable disap- 
pears, consider: 

1. If the resulting statement is false, then you have two distinct parallel lines 
and there is no solution. 

2. If the resulting statement is true, then you have the same lines and there 
are an infinite number of solutions. 



270 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



f> t* t* Exercises ■** •** ■** 

In Exercises 1-8, use the substitution method to solve each of the following systems. Check your answer 
manually, without the use of a calculator. 

1. — 7a: + 7y = 63 5. x = —5 — 2y 

y = 6-2x -2x-Gy = 18 

2. 3a;-8y = 27 6. x = 15 + 6y 

y = 4 - 7x 9x + 3y = 21 

3. x = 19 + 7y 7. 6a; - 8y = 24 

3a: — 3y = 3 y = 15 + 3a; 

4. a; = 39 + 8y 8. 9a; + 8y = -45 
-9a; + 2y = -71 y = 15 - 8a; 



In Exercises 9-28, use the substitution method to solve each of the following systems. 

18. -6x-6y = 102 
9x — y = —63 

10. -a; + 9y = -12 19. -Ax - 8y = -4 

y = -2j/-4 

11. -x + 4y = 22 20. 3x + 6y = 2 

y = -2y + 2 

12. -x-2y=15 21. -2a;-2y = 26 

-7a; + y = 19 

13. x + 2y = -A 22. -2x - 8y = -30 



— x + 9y = 


= 46 


7x — Ay = 


= -27 


—x + 9y = 


= -12 


Ax + 7y = 


= -38 


—x + Ay = 


= 22 


8x + 7y = 


= -20 


-x-2y = 


= 15 


3a; — 9y = 


= 15 


x + 2y = 


-4 


6x — Ay = 


-56 


x + 8y = 


79 


Ax + Qy = 


82 


x + 6y 


= -49 


-3x + Ay 


= -7 


x — Ay = 


33 


Ax + 7y = 


-6 


-2x + 8y 


= -50 


—9x — y 


= -3 



—6a; + y = 10 

14. a; + 8y = 79 23. 3a; - Ay = -43 

-3x + y = 22 

15. a; + 6y=-49 24. -2x + 8y = 14 

8x + y = A3 

16. a;-4y = 33 25. -8a; - Ay = 24 

9x-y = -71 

17. -2a; + 8y=-50 26. -8x - 2y = -1A 

—6a; — y = —9 



4.2. SOLVING SYSTEMS BY SUBSTITUTION 271 



27. -8x-7y = 2 28. 9x + 4y = -3 

8 9 

y = — —x + 9 y = ——x + o 



In Exercises 29-36, use the substitution method to solve each of the following systems. Use your 
graphing calculator to check your solution. 

29. 3x-5y = 3 33. 3x + 8y = 6 

5x - 7y = 2 2x + 7y = -2 

30. 4a; - 5y = 4 34. 3a; - 7y = 6 

3a; — 2y = — 1 2a; — 3y = 1 

31. 4x + 3y = 8 35. 4a; + 5y = 4 
3a; + 4 y = 2 -3x - 2y = 1 

32. 3a; + 8y = 3 36. 5a; + Ay = 5 

-4a; - 9y = -2 4a; + 5y = 2 



In Exercises 37-48, use the substitution method to determine how many solutions each of the following 
linear systems has. 

37. -9a; + 6y = 9 43. y = 7y + 18 

9a; - 63y = 162 

44. y = Ay - 9 

-10a; + 4Ch/ = 90 



39. 



40. 



-9a; + 6y = 9 




3 

V = —x — 8 
y 2 




3a; — 5y = 9 




3 ^fi 
u = -x + 

5 




y = -2a;- 


16 


-14a;- 7y = 112 




y = -12a;- 


f 12 


120a;+102/= 120 




x = 16 — 5y 




-Ax + 2y = 2A 




x = -18- Ay 




7x-7y = 49 





45. 



46. 



a;= -2y + 3 

Ax + 8y = A 

x = 2y + A 
—3x + 6y = 5 



41. a;=16-5w 47 - -9a; + 4y = 73 

y = —3 — 2x 

42. a;=-18-4y 48. 6x + 9y = 27 

y = 16 — 5a; 



1. 


-1,8) 


3. 


-2,-3) 


5. 


3,-4) 


7. 


-8,-9) 


9. 


-1,5) 


11. 


-6,4) 


13. 


-8,2) 


15. 


-7,-7) 


17. 


1,-6) 


19. 




21. 


-4,-9) 


23. 


-5,7) 



272 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



£»- j*- &*■ Answers •*$ **s ■*$ 

25. (-7,8) 

27. No solution 

29. (-11/4,-9/4) 

31. (26/7,-16/7) 

33. (58/5,-18/5) 

35. (-13/7,16/7) 

37. No solutions 

39. Infinite number of solutions 

41. One solution 

43. Infinite number of solutions 

45. No solutions 

47. One solution 



4.3. SOLVING SYSTEMS BY ELIMINATION 



273 



4.3 Solving Systems by Elimination 

When both equations of a system are in standard form Ax + By = C, then 
a process called elimination is usually the best procedure to use to find the 
solution of the system. Elimination is based on two simple ideas, the first of 
which should be familiar. 

1. Multiplying both sides of an equation by a non-zero number does not 
change its solutions. Thus, the equation 



x + 3y = 7 



(4.19) 



will have the same solutions (it's the same line) as the equation obtained 
by multiplying equation (4.19) by 2. 



2x + 6y = 14 



(4.20) 



2. Adding two true equations produces another true equation. For example, 
consider what happens when you add 4 = 4 to 5 = 5. 



Even more importantly, consider what happens when you add two equa- 
tions that have (2, 1) as a solution. The result is a third equation whose 
graph also passes through the solution. 



X 


+ 


V 


= 


3 


X 


- 


y 


= 


1 


2x 






= 


4 


X 






= 


2 




> X 



Fact. Adding a multiple of an equation to a second equation produces and 
equation that passes through the same solution as the first two equations. 



274 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



One more important thing to notice is the fact that when we added the 
equations 

x + y = 3 

x - y = I 

2x =4 

the variable y was eliminated. This is where the elimination method gets its 
name. The strategy is to somehow add the equations of a system with the 
intent of eliminating one of the unknown variables. 

However, sometimes you need to do a little bit more than simply add the 
equations. Let's look at an example. 



You Try It! 



Solve the following system of EXAMPLE 1. Solve the following system of equations, 
equations: 

x + 2y = -5 

2x — y = —5 



x + 3y = 14 
-8x-3y = -28 



(4.21) 
(4.22) 



Solution: Our focus will be on eliminating the variable x. Note that if we 
multiply equation (4.21) by —2, then add the result to equation (4.22), the x 
terms will be eliminated. 



-2x 
2x 



v 



10 
-5 



% 



Multiply equation (4.21) by -2. 
Equation (4.22). 

Add the equations. 













I 

r > 


'2s- 


V 


= 


-5 














7 
























/ 






















/ 
























/ 






















/ 




















£ 


7 


k 


-3 


j 


1) 








c 




















/ 






















/ 
























/ 








5- 















Divide both sides of — by = 5 by — 5 to get y = — 1. 

To find the corresponding value of x, substitute —1 for y in equation (4.21) 
(or equation (4.22)) and solve for x. 



Equation (4.21) 
Substitute —1 for y. 
Solve for x. 



x + 2y = — 5 Check: To check, we need to show that the point (x, y) = (—3,1) satisfies 

both equations. 
Figure 4.38: x + 2y = -5 

and 2x - y = -5 intersect at Substitute (x,y) = (-3,-1) into Substitute (x,y) = (-3,-1) into 

( — 3,— 1). equation (4.21). equation (4.22). 



x + 2y = 


-5 


+ 2(-l) = 


-5 


x = 


-3 



x + 2y 

-3 + 2(-l) 

-5 







2x- 


-y = 


-5 


2(" 


-3) 


"(- 


-5 = 


-5 
-5 



Answer: (2,4). 



Thus, the point (x, y) = (—3, —1) satisfies both equations and is therefore the 
solution of the system. 



□ 



4.3. SOLVING SYSTEMS BY ELIMINATION 



275 



To show that you have the option of which variable you choose to eliminate, 
let's try Example 1 a second time, this time eliminating y instead of x. 



EXAMPLE 2. Solve the following system of equations. 

x + 2y = -5 (4.23) 

2x-y=-5 (4.24) 

Solution: This time we focus on eliminating the variable y. We note that if 
we multiply equation (4.24) by 2, then add the result to equation (4.23), the y 
terms will be eliminated. 



x 
Ax 



2y 
2y 



5.7,' 



-5 Equation (4.23). 

-10 Multiply equation (4.24) by 2. 

-15 Add the equations. 



Divide both sides of 5a; = —15 by 5 to get x = — 3. 

To find the corresponding value of y, substitute —3 for x in equation (4.23) 
(or equation (4.24)) and solve for y. 



x 
-3 



•2» 

2y 

2y 

y 



Equation (4.23) 
Substitute —3 for x. 
Add 3 to both sides. 
Divide both sides by 2. 



Hence, (x, y) = (—3, —1), just as in Example 1, is the solution of the system. 



Sometimes elimination requires a thought process similar to that of finding 
a common denominator. 



EXAMPLE 3. Solve the following system of equations. 

3x + Ay = 12 
2x - 5y = 10 



(4.25) 
(4.26) 



Solution: Let's focus on eliminating the x-terms. Note that if we multiply 
equation (4.25) by 2, then multiply equation (4.26) by —3, the x-terms will be 
eliminated when we add the resulting equations. 



6.7' 

-Qx 



8y 
15y 



24 
-30 



23y 



Multiply equation (4.25) by 2. 
Multiply equation (4.26) by —3. 

Add the equations. 



You Try It! 




Figure 4.39: x + 2y = -5 
and 2x — y = —5 intersect at 

(-3,-1). 



□ 



You Try It! 



Solve the following system of 
equations: 



-I4x + 9y 
7x + 3y 



94 



-62 



276 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Hence, y = -6/23. 

At this point, we could substitute y = —6/23 in either equation, then solve 
the result for x. However, working with y = —6/23 is a bit daunting, partic- 
ularly in the light of elimination being easier. So let's use elimination again, 
this time focusing on eliminating y. Note that if we multiply equation (4.25) 
by 5, then multiply equation (4.26) by 4, when we add the results, the y-terms 
will be eliminated. 




12 



Figure 4.40: 3a; + 4y = 12 
and 2x — 5y = 10 intersect at 
(100/23,-6/23). 



Answer: 



15a; 

8x 



20y 
20y 



60 
40 



23a: 
Thus, x 
Check: 



Multiply equation (4.25) by 5. 
Multiply equation (4.26) by 4. 

= 100 Add the equations. 

100/23, and the system of the system is (x,y) = (100/23, 



-6/23). 



Let's use the graphing calculator to check the solution. First, store 
100/23 in X, then —6/23 in Y (see Figure 4.41). Next, enter the left-hand sides 
of equations (4.25) and (4.26). 



4.347826087 
-6^23+V 

-.2608695652 



3*X+4*V 
2*X-5*Y 



12 
10 



Figure 4.41: Enter 100/23 in X, 
-6/23 in Y. 



Figure 4.42: Enter the left-hand 
sides of equations (4.25) and (4.26). 



Note that both calculations in Figure 4.42 provide the correct right-hand sides 
for equations (4.25) and (4.26). Thus, the solution (x,y) = (100/23,-6/23) 
checks. 

□ 



You Try It! 



Exceptional Cases 

In the previous section, we saw that if the substitution method led to a false 
statement, then we have parallel lines. The same thing can happen with the 
elimination method of this section. 



Solve the following system of EXAMPLE 4. Solve the following system of equations, 
equations: 



5a; — Ay 
15a; - 12y 



16 



x + y 
2x + 2y 



-6 



(4.27) 
(4.28) 



49 



4.3. SOLVING SYSTEMS BY ELIMINATION 



277 



Solution: Let's focus on eliminating the x-terms. Note that if we multiply 
equation (4.27) by —2, the x-terms will be eliminated when we add the resulting 
equations. 

-2x - 2y = -6 Multiply equation (4.27) by -2. 

2x + 2y = -6 Equation (4.28). 

= —12 Add the equations. 



Because of our experience with this solving this exceptional case with substi- 
tution, the fact that both variables have disappeared should not be completely 
surprising. Note that this last statement is false, regardless of the values of x 
and y. Hence, the system has no solution. 

Indeed, if we find the intercepts of each equation and plot them, then we can 
easily see that the lines of this system are parallel (see Figure 4.43). Parallel 
lines never intersect, so the system has no solutions. 




x + y = 3 



2x + 2y = -6 
Figure 4.43: The lines x + y = 3 and 2x + 2y = —6 are parallel. 



Answer: no solution 



In the previous section, we saw that if the substitution method led to a true 
statement, then we have the same lines. The same thing can happen with the 
elimination method of this section. 



□ 



EXAMPLE 5. Solve the following system of equations. 

x — 7y = 4 
-3x + 2ly = -12 







You Try It! 












Solve t 


le following sys 


;em of 


(4.29) 


equations: 


(4.30) 


2x - ly = 4 






8a; - 28y = 16 





278 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 




Solution: If we are not on automatic pilot late at night doing our homework, 
we might recognize that the equations 4.29 and (4.30) are identical. But it's 
also conceivable that we don't see that right away and begin the elimination 
4 method. Let's multiply the first equation by 3, then add. This will eliminate 
the x-terms. 

3a: - 21y = 12 Multiply equation (4.29) by 3. 

-3a; + 21y = -12 Equation (4.30). 

= Add the equations. 

Fieure 4 44- x — 7v = 4 and Again, all of the variables have disappeared! However, this time the last state- 

o„ , oi v — _io are the men t is true, regardless of the values of x and y. 
same line Infinite number of Notice that if we multiply equation (4.29) by —3, then we have two identical 

solutions. equations. 



Answer: There are an 
infinite number of solutions. 
Examples of solution points 
are (2,0), (9,2), and 
(-5,-2). 



-3a; + 2ly = -12 Multiply equation (4.29) by 3. 

-3a; + 21y = -12 Equation (4.30). 

The equations are identical! Hence, there are an infinite number of points of 
intersection. Indeed, any point on either line is a solution. Example points of 
solution are (-3,-1), (0,-4/7), and (4,0). 

□ 



4.3. SOLVING SYSTEMS BY ELIMINATION 



279 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



In Exercises 1-8, use the elimination method to solve each of the following systems. Check your result 
manually, without the assistance of a calculator. 



x + Ay 



9x 



7.y 




-43 



x + 6y = 


-53 


5a; — 9y = 


47 


6a; + 2/ = 


8 


4a; + 2y = 





Ax + y = 


18 


— 2x + 6y = 


= -22 



7. 



-8a; + y 
Ax + 3y 


= -56 
56 


2x + y = 
7x + 8y = 


= 21 

= 87 


x + 8y 
— bx — 9y 


41 
= -50 


x — Ay 
—2x — 6y 


= -31 
= -36 



In Exercises 9-16, use the elimination method to solve each of the following systems. 



10. 



11. 



12. 



-12a; + 9y = 





—6a; — Ay = 


-34 


-27a; - by = 


148 


— 9x — 3y = 


60 


27a; — 6y = 


-96 


-3x — by = 


22 


-8x + 8y = 


-32 


2x - 9y = 


15 



13. 



14. 



15. 



16. 



2x 


- 


Qy 


28 


3x 


+ 


I8y 


= -60 


8x 


— 


6y 


96 


Ax 


+ 


30y 


= -156 


32a; 


+ 


111 


= -238 


8a: 


- 


Mj 


64 


2x 


+ 


Qy 


= 30 


2x 


+ 


7y 


= 51 



In Exercises 17-24, use the elimination method to solve each of the following systems. 



17. 



18. 



19. 



20. 



3.r 


- 7y = 


-75 


-2a; 


- 2y = 


-10 


-8a; 


+ 3y = 


42 


-7x 


+ 8y = 


20 


9x - 


- 9y = 


-63 


2x - 


- Qy = 


-34 


-Ax 


- 8y = 


-52 


-7x 


- 3y = 


-14 



21. 



22. 



23. 



24. 



9;r 


- 


2y = 


28 


5 a: 


- 


3y = 


= -32 


8a; 


— 


2y = 


= -12 


Gx 


+ 


3y = 


12 


3x 


— 


by = 


= -34 


7x 


+ 


7y ~- 


56 


9x 


— 


9y = 


= 9 


7x 


+ 


Ay = 


= 8 



280 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



In Exercises 25-32, use the elimination method to solve each of the following systems. Use your 
calculator to check your solutions. 



25. 



26. 



27. 



28. 



2a; - 

7x + 


7y = 
6y = 


-2 
3 


-9x - 
bx — 


Ay = 
3y = 


4 
-1 


2x + 
-5x + 


3y = 

by = 


-2 

2 


-bx + 
-Ax - 


8y = 
7y = 


-3 
3 



29. 



30. 



31. 



32. 



9a; + Ay 
— 7x — 9y 


= -A 
3 


— 3a; — by 

Ax + 6y 


= -4 
= 1 


2x + 2y = 
3a; — by - 


= 4 
= 3 


6a; — 9y 

-Ax - %y 


= -2 
4 



In Exercises 33-40, use the elimination method to determine how many solutions each of the following 
system of equations has. 



33. 



34. 



35. 



36. 



X 


+ 


7y = 


-32 


-8a; 


- 


56y = 


256 


-8a; 


+ 


y = 


-53 


56a; 


- 


7y = 


371 


16a; 


— 


16i/ = 


-256 


-8a; 


+ 


8y = 


128 


3.r 


— 


3y = 


42 


-6a; 


+ 


6y = 


-84 



37. 



38. 



39. 



40. 



X 


- Ay = 


-37 


2x 


- 8y = 


54 


Ax 


+ y = 


-13 


28a; 


+ 7y = 


189 


X 


+ 9y = 


73 


-Ax 


- by = 


-44 


O.r 


+ y = 


31 


— 5a; 


- 6y = 


-62 



j*- j*- ?*- Answers •** ~*^ ■** 



1. (-4,1) 
3. (2,-4) 
5. (8,8) 
7.(1,5) 
9. (3,4) 
11. (-4,-2) 



13. 


(8,-2) 


15. 


(7,-2) 


17. 


(-4,9) 


19. 


(-2,5) 


21. 


(-4,4) 


23. 


(3,5) 



4.3. SOLVING SYSTEMS BY ELIMINATION 281 

25. (9/61, 20/61) 33. Infinite number of solutions 

27. (—16/25, —6/25) 35. Infinite number of solutions 

29. (-24/53, 1/53) 37. No solutions 

31. (13/8,3/8) 39. One solution 



282 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



You Try It! 



If the second of two 
complementary angles is 6 
degrees larger than 3 times 
the first angle, find the 
degree measure of both 
angles. 



4.4 Applications of Linear Systems 

In this section we create and solve applications that lead to systems of linear 
equations. As we create and solve our models, we'll follow the Requirements for 
Word Problem Solutions from Chapter 2, Section 5. However, instead of setting 
up a single equation, we set up a system of equations for each application. 



EXAMPLE 1. In geometry, two angles that sum to 90° are called comple- 
mentary angles. If the second of two complementary angles is 30 degrees larger 
than twice the first angle, find the degree measure of both angles. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Our variable dictionary will take the form 
of a diagram, naming the two complementary angles a and /3. 




2. Set up a Systems of Equations. The "second angle is 30 degrees larger 
than twice the first angle" becomes 



P = 30 + 2a 



(4.31) 



Secondly, the angles are complementary, meaning that the sum of the 
angles is 90°. 

a + /3 = 90 (4.32) 

Thus, we have a system of two equations in two unknowns a and (3. 

3. Solve the System. As equation (4.31) is already solved for /?, let use the 
substitution method and substitute 30 + 2a for j3 in equation (4.32). 



a + (3 = 90 

a + (30 + 2a) = 90 

3a + 30 = 90 

3a = 60 

a = 20 



Equation (4.32). 
Substitute 30 + 2a for (3. 
Combine like terms. 
Subtract 30 from both sides. 
Divide both sides by 3. 



4.4. APPLICATIONS OF LINEAR SYSTEMS 



283 



4. Answer the Question. The first angle is a = 20 degrees. The second 
angle is: 



f3 = 30 + 2a 
/3 = 30 + 2(20) 
/3 = 70 



Equation (4.31). 
Substitute 20 for a. 
Simplify. 



5. Look Back. Certainly 70° is 30° larger than twice 20°. Also, note that 
20° + 70° = 90°, so the angles are complementary. We have the correct 
solution. 



Answer: 21 and 69 



□ 



EXAMPLE 2. The perimeter of a rectangle is 280 feet. The length of the 
rectangle is 10 feet less than twice the width. Find the width and length of the 
rectangle. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Our variable dictionary will take the form 
of a diagram, naming the width and length W and L, respectively. 



You Try It! 



The perimeter of a rectangle 
is 368 meters. The length of 
the rectangle is 34 meters 
more than twice the width. 
Find the width and length of 
the rectangle. 



W 



L 



W 



2. Set up a System of Equations. The perimeter is found by summing the 
four sides of the rectangle. 

P=L+W +L+W 
P = 2L + 2W 

We're told the perimeter is 280 feet, so we can substitute 280 for P in 
the last equation. 



280 = 2L + 2W 



284 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 

We can simplify this equation by dividing both sides by 2, giving the 
following result: 

L + W = 140 

Secondly, we're told that the "length is 10 feet less than twice the width." 
This translates to: 

L = 2W - 10 

Thus, the system we need to solve is: 

L + W = U0 (4.33) 

L = 2W-W (4.34) 

3. Solve the System. As equation (4.34) is already solved for L, let use the 
substitution method and substitute 2W — 10 for L in equation (4.33). 

W + L = 140 Equation (4.33). 

W + (2W - 10) = 140 Substitute 2W - 10 for L. 

3W - 10 = 140 Combine like terms. 

3W = 150 Add 10 to both sides. 

W = 50 Divide both sides by 3. 

4. Answer the Question. The width is W = 50 feet. The length is: 

L = 2W-W Equation (4.34). 

L = 2(50) - 10 Substitute 50 for W. 

L = 90 Simplify. 

Thus, the length is L = 90 feet. 

5. Look Back. Perhaps a picture, labeled with our answers might best 
demonstrate that we have the correct solution. Remember, we found 
that the width was 50 feet and the length was 90 feet. 

90 



50 



50 



Answer: length = 134, 
width = 50 



90 



Note that the perimeter is P = 90 + 50 + 90 + 50 = 280 feet. Secondly, 
note that the length 90 feet is 10 feet less than twice the width. So we 
have the correct solution. 



□ 



4.4. APPLICATIONS OF LINEAR SYSTEMS 



285 



You Try It! 



EXAMPLE 3. Pascal has $3.25 in change in his pocket, all in dimes and Eloise has $7.10 in change in 
quarters. He has 22 coins in all. How many dimes does he have? her pocket, all in nickels and 

quarters, she has 46 coins in 
all. How many quarters does 
she have? 



Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 



1 . Set up a Variable Dictionary. Let D represent the number of dimes and 
let Q represent the number of quarters. 

2. Set up a System of Equations. Using a table to summarize information is 
a good strategy. In the first column, we list the type of coin. The second 
column gives the number of each type of coin, and the third column 
contains the value (in cents) of the number of coins in Pascal's pocket. 



Number of Coins Value (in cents) 


Dimes D WD 


Quarters Q 25Q 


Totals 22 325 



Note that D times, valued at 10 cents apiece, are worth 10D cents. Sim- 
ilarly, Q quarters, valued at 25 cents apiece, are worth 25Q cents. Note 
also how we've change $3.25 to 325 cents. 

The second column of the table gives us our first equation. 

D + Q = 22 (4.35) 

The third column of the table gives us our second equation. 

10£> + 25Q = 325 (4.36) 



Solve the System. Because equations (4.35) and (4.36) are both in stan- 
dard form Ax + By = C, we'll use the elimination method to find a 
solution. Because the question asks us to find the number of dimes in 
Pascal's pocket, we'll focus on eliminating the Q-terms and keeping the 
D-terms. 



-25D 
10D 



25Q 
25Q 



-15D 



-550 Multiply equation (4.35) by -25. 

325 Equation (4.36). 

-225 Add the equations. 



Dividing both sides of the last equation by —15 gives us D = 15. 

4. Answer the Question. The previous solution tells us that Pascal has 15 
dimes in his pocket. 



286 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Answer: 24 



5. Look Back. Again, summarizing results in a table might help us see if we 
have the correct solution. First, because we're told that Pascal has 22 
coins in all, and we found that he had 15 dimes, this means that he must 
have 7 quarters. 



Number of Coins Value (in cents) 


Dimes 15 150 


Quarters 7 175 


Totals 22 325 



Fifteen dimes are worth 150 cents, and 7 quarters are worth 175 cents. 
That's a total of 22 coins and 325 cents, or $3.25. Thus we have the 
correct solution. 



□ 



You Try It! 



Eileen inherits $40,000 and 
decides to invest the money 
in two accounts, part in a 
certificate of deposit that 
pays 3% interest per year, 
and the rest in a mutual 
fund that pays 6% per year. 
At the end of the first year, 
her investments earn a total 
of $2,010 in interest. Find 
the amount invested in each 
account. 



EXAMPLE 4. Rosa inherits $10,000 and decides to invest the money in 
two accounts, one portion in a certificate of deposit that pays 4% interest per 
year, and the rest in a mutual fund that pays 5% per year. At the end of the 
first year, Rosa's investments earn a total of $420 in interest. Find the amount 
invested in each account. 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Let C represent the amount invested in the 
certificate of deposit and M represent the amount invested in the mutual 
fund. 

2. Set up a System of Equations. We'll again use a table to summarize 
information. 



Rate Amount invested Interest 


Certificate of Deposit 4% C 0.04C 


Mutual Fund 5% M 0.05M 


Totals 10,000 420 



At 4%, the interest earned on a C dollars investment is found by taking 
4% of C (i.e., 0.04C). Similarly, the interest earned on the mutual fund 
is 0.05M. 



4.4. APPLICATIONS OF LINEAR SYSTEMS 287 

The third column of the table gives us our first equation. The total 
investment is $10,000. 

C + M = 10000 

The fourth column of the table gives us our second equation. The total 
interest earned is the sum of the interest earned in each account. 

0.04(7 + 0.05M = 420 

Let's clear the decimals from the last equation by multiplying both sides 
of the equation by 100. 

4(7 + 5M = 42000 

Thus, the system we need to solve is: 

C + M = 10000 (4.37) 

4(7 + 5M = 42000 (4.38) 

3. Solve the System. Because equations (4.37) and (4.38) are both in stan- 
dard form Ax + By = C, we'll use the elimination method to find a 
solution. We'll focus on eliminating the (7-terms. 

-AC - AM = -40000 Multiply equation (4.37) by -4. 

AC + 5M = 42000 Equation (4.38). 



M = 2000 Add the equations. 

Thus, the amount invested in the mutual fund in M = $2, 000. 

4. Answer the Question. The question asks us to find the amount invested 
in each account. So, substitute M = 2000 in equation (4.37) and solve 
for (7. 

C + M = 10000 Equation (4.37). 

C + 2000 = 10000 Substitute 2000 for M. 

C = 8000 Subtract 2000 from both sides. 

Thus C = $8, 000 was invested in the certificate of deposit. 

5. Look Back. First, note that the investments in the certificate of deposit 
and the mutual fund, $8,000 and $2,000 respectively, total $10,000. Let's 
calculate the interest on each investment: 4% of $8,000 is $320 and 5% 
of $2,000 is $100. 



Rate Amount invested Interest 


Certificate of Deposit 4% 8,000 320 


Mutual Fund 5% 2, 000 100 


Totals 10,000 420 



288 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



Answer: $13,000 in the 
certificate of deposit, $27,000 
in the mutual fund. 



Note that the total interest is $420, as required in the problem statement. 
Thus, our solution is correct. 



□ 



You Try It! 



A store sells peanuts for 
$4.00 per pound and pecans 
for $7.00 per pound. How 
many pounds of peanuts and 
how many pounds of pecans 
should you mix to make a 
25-lb mixture costing $5.80 
per pound? 



EXAMPLE 5. Peanuts retail at $0.50 per pound and cashews cost $1.25 per 
pound. If you were a shop owner, how many pounds of peanuts and cashews 
should you mix to make 50 pounds of a peanut-cashew mixture costing $0.95 
per pound? 

Solution: In the solution, we address each step of the Requirements for Word 
Problem Solutions. 

1. Set up a Variable Dictionary. Let P be the number of pounds of peanuts 
used and let C be the number of pounds of cashews used. 

2. Set up a System of Equations. We'll again use a table to summarize 
information. 



Cost per pound Amount (pounds) Cost 


Peanuts $0.50 P 0.50P 


Cashews $1.25 C 1.25C 


Totals $0.95 50 0.95(50)=47.50 



At $0.50 per pound, P pounds of peanuts cost 0.50P. At $1.25 per pound, 
C pounds of cashews cost 1.25C. Finally, at $0.95 per pound, 50 pounds 
of a mixture of peanuts and cashews will cost 0.95(50), or $47.50. 

The third column of the table gives us our first equation. The total 
number of pounds of mixture is given by the following equation: 

P + C = 50 

The fourth column of the table gives us our second equation. The total 
cost is the sum of the costs for purchasing the peanuts and cashews. 

0.50P + 1.25C = 47.50 

Let's clear the decimals from the last equation by multiplying both sides 
of the equation by 100. 

50P + 125C = 4750 

Thus, the system we need to solve is: 

P + C = 50 (4.39) 

50P + 125C = 4750 (4.40) 



4.4. APPLICATIONS OF LINEAR SYSTEMS 



289 



3. Solve the System. Because equations (4.39) and (4.40) are both in stan- 
dard form Ax + By = C, we'll use the elimination method to find a 
solution. We'll focus on eliminating the P-terms. 



-50P 
50P 



50C 
125C* 



-2500 
4750 



Multiply equation (4.39) by -50. 
Equation (4.40). 

75 C = 2250 Add the equations. 

Divide both sides by 75 to get C = 30 pounds of cashews are in the mix. 

Answer the Question. The question asks for both amounts, peanuts and 
cashews. Substitute C = 30 in equation (4.39) to determine P. 



P 
P - 



-C 
30 
P 



50 
50 
20 



Equation (4.39). 
Substitute 30 for C. 
Subtract 30 from both sides. 



Thus, there are P = 20 pounds of peanuts in the mix. 

Look Back. First, note that the amount of peanuts and cashews in the 
mix is 20 and 30 pounds respectively, so the total mixture weighs 50 
pounds as required. Let's calculate the costs: for the peanuts, 0.50(20), 
or $10, for the cashews, 1.25(30) = 37.50. 



Cost per pound Amount (pounds) Cost 


Peanuts $0.50 20 $10.00 


Cashews $1.25 30 $37.50 


Totals $0.95 50 47.50 



Note that the total cost is $47.50, as required in the problem statement. 
Thus, our solution is correct. 



Answer: 10 pounds of 
peanuts, 15 pounds of 
pecans 



□ 



290 



CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 



f»- n- fa- 



Exercises 



■*1 ■*'» .*5 



1. In geometry, two angles that sum to 90° 
are called complementary angles. If the 
second of two complementary angles is 42 
degrees larger than 3 times the first angle, 
find the degree measure of both angles. 

2. In geometry, two angles that sum to 90° 
are called complementary angles. If the 
second of two complementary angles is 57 
degrees larger than 2 times the first angle, 
find the degree measure of both angles. 

3. The perimeter of a rectangle is 116 inches. 
The length of the rectangle is 28 inches 
more than twice the width. Find the 
width and length of the rectangle. 

4. The perimeter of a rectangle is 528 inches. 
The length of the rectangle is 24 inches 
more than twice the width. Find the 
width and length of the rectangle. 

5. Maria has $6.35 in change in her pocket, 
all in nickels and quarters, she has 59 
coins in all. How many quarters does she 
have? 

6. Amy has $5.05 in change in her pocket, all 
in nickels and quarters, she has 53 coins 
in all. How many quarters does she have? 

7. A store sells cashews for $6.00 per pound 
and raisins for $7.00 per pound. How 
many pounds of cashews and how many 
pounds of raisins should you mix to make 
a 50-lb mixture costing $6.42 per pound? 

8. A store sells cashews for $3.00 per pound 
and pecans for $8.00 per pound. How 
many pounds of cashews and how many 
pounds of pecans should you mix to make 
a 50-lb mixture costing $4.10 per pound? 

9. Roberto has $5.45 in change in his pocket, 
all in dimes and quarters, he has 38 coins 
in all. How many dimes does he have? 



10. Benjamin has $7.40 in change in his 
pocket, all in dimes and quarters, he has 
44 coins in all. How many dimes does he 
have? 

11. In geometry, two angles that sum to 180° 
are called supplementary angles. If the 
second of two supplementary angles is 40 
degrees larger than 3 times the first angle, 
find the degree measure of both angles. 

12. In geometry, two angles that sum to 180° 
are called supplementary angles. If the 
second of two supplementary angles is 114 
degrees larger than 2 times the first angle, 
find the degree measure of both angles. 

13. Eileen inherits $20,000 and decides to in- 
vest the money in two accounts, part in 
a certificate of deposit that pays 3% in- 
terest per year, and the rest in a mutual 
fund that pays 5% per year. At the end 
of the first year, her investments earn a 
total of $780 in interest. Find the amount 
invested in each account. 

14. Alice inherits $40,000 and decides to in- 
vest the money in two accounts, part in 
a certificate of deposit that pays 3% in- 
terest per year, and the rest in a mutual 
fund that pays 6% per year. At the end of 
the first year, her investments earn a to- 
tal of $1,980 in interest. Find the amount 
invested in each account. 

15. The perimeter of a rectangle is 376 cen- 
timeters. The length of the rectangle is 
12 centimeters less than three times the 
width. Find the width and length of the 
rectangle. 

16. The perimeter of a rectangle is 344 feet. 
The length of the rectangle is 28 feet less 
than three times the width. Find the 
width and length of the rectangle. 



4.4. APPLICATIONS OF LINEAR SYSTEMS 



291 



$*•$*<$*■ Answers •*$ *** •** 



1. 12° and 78° 

3. Length is 48 inches, width is 10 inches 

5. 17 quarters 

7. 29 pounds of cashews, 21 pounds of raisins 

9. 27 dimes 



11. 35° and 145° 

13. $11,000 in certificate of deposit, $9,000 in 
mutual fund. 

15. Length is 138 centimeters, width is 50 cen- 
timeters 



Chapter 5 



Polynomial Functions 



Polynomials appear in a wide variety of applications in mathematics and sci- 
ence. Solving algebraic equations containing polynomials is among the oldest 
problems in mathematics. The Babylonians solved quadratic equations as early 
as 2000 BC. Euclid of Alexandria used geometry to solve quadratic equations 
in 300 BC. Arab mathematicians solved quadratic equations in 1000 AD. The 
current method of solving equations using more modern notation started in 
1557 when Robert Recorde used the equal sign in The Whetstone of Witte. 

In this chapter, we will learn about the properties of polynomials, as well 
as how to manipulate and use them in application problems. 



293 



294 CHAPTER 5. POLYNOMIAL FUNCTIONS 

5.1 Functions 

We begin with the definition of a relation. 

Relation. A relation is a collection of ordered pairs. 

The collection of ordered pairs 

R = {(0,3), (0,4), (1,5), (2, 6)} 

is an example of a relation. 

If we collect the first element of each ordered pair in a set, we have what is 
called the domain of the relation. 

Domain. The domain of a relation is the set of all first elements of the ordered 
pairs. 

For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the 
first element of each ordered pair in i?, we get the domain: 

Domain of R = {0,1,2} 

Although the number zero appears twice as a first element in the ordered pairs 
of R, note that we list it only once when listing the elements in the domain of 
R. 

In similar fashion, if we collect the second elements of each ordered pair in 
a set, we have what is called the range of a relation. 

Range. The range of a relation is the set of all second elements of the ordered 
pairs. 

For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the 
second element of each ordered pair in i?, we get the range: 

Range of R= {3,4,5,6} 



You Try It! 



State the domain and range EXAMPLE 1. State the domain and range of the relation 

of the relation 

T = {(5,3), (6,3), (7,4)}. 

S = {(-1,7), (2,5), (2,3)}. 

Solution: Collect the first element of each ordered pair in T to list the 
domain: 

Domain of T = {5, 6, 7} 



5.1. FUNCTIONS 



295 



Collect the second element of each ordered pair in T to list the range: 

Range of T = {3, 4} 

Note that even though the number three appears in the second position twice, Answer: 
we list it only once in describing the range. Domain of S = { — 1, 2}, 

Range of S = {3, 5, 7} 



□ 



You Try It! 



EXAMPLE 2. State the domain and range of the relation shown in Figure 5.1. State the domain and range 

of the relation shown below. 

y 
y 



C 



B 



D 



■*■ x 











5' 


























L 










J 




























































5 


















*, 




A 




































































5 > 


■ h 











Figure 5.1: We can present a relation as a collection of ordered pairs in a graph. 



Solution: Point A has coordinates (3,2), point B has coordinates (—2,3), 
point C has coordinates (—4,-3), and point D has coordinates (1,-3). We 
can collect these points in a set. 

5 = {(3,2), (-2,3), (-4,-3), (1,-3)} 

If we collect each element in the first position of each ordered pair, we have the 
domain. 

Domain of S = {-4, -2, 1,3} 

Note that it is traditional to list the domain elements in order (smallest to 
largest). Next, if we collect each element in the second position of each ordered 
pair, we have the range. 



Range of S= {-3,2,3} 



296 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Answer: 

Domain of 5 = {-2,1,2}, 

Range of 5 = {-4,-1,2,3} 



Again, it is traditional to list the elements in order. Note again that we 
did not repeat the number —3 in listing the range, even though it is used twice 
as a second element of an ordered pair in the set S. 

□ 



Mapping Diagrams 

A mapping diagram is a useful construction that helps one to analyze a relation. 
Consider the earlier relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, which had domain 
V = {0, 1, 2} and range TZ = {3, 4, 5, 6}. To construct a mapping diagram for 
R, list the elements in the domain of R on the left, list the elements of the 
range of R on the right, then use arrows to indicate the ordered pairs (see 
Figure 5.2). 



R 




Figure 5.2: Using a mapping diagram to describe the relation R. 

Note how the ordred pair (0, 3) is indicated by drawing an arrow connecting 
on the left to 3 on the right. We say that the relation "maps to 3" and write 
R : — > 3. In similar fashion: 

• The ordered pair (0, 4) is indicated by drawing an arrow connecting on 
the left to 4 on the right; i.e., R "maps to 4" or R : — > 4. 

• The ordered pair (1,5) is indicated by drawing an arrow connecting 1 on 
the left to 5 on the right; i.e., R "maps 1 to 5" or R : 1 — > 5. 

• The ordered pair (2, 6) is indicated by drawing an arrow connecting 2 on 
the left to 6 on the right; i.e., R "maps 2 to 6" or R : 2 — > 6. 



You Try It! 



EXAMPLE 3. Create a mapping diagram for the relation in Example 1. 

Solution. The relation of Example 1 is T = {(5, 3), (6, 3), (7, 4)}. List the 
domain T> = {5, 6, 7} on the left, the range TZ = {3, 4} on the right, then use 
arrows to indicate the ordered pairs (see Figure 5.3). 

□ 



5.1. FUNCTIONS 



297 




Figure 5.3: A mapping diagram for the relation T. 

Function Definition 

A function is a very special type of relation. 



Function A relation is a function if and only if each object in the domain is 
paired with exactly one object in the range. 



As a first example, consider the relation R = {(0,3), (0,4), (1,5), (2,6)} whose 
mapping diagram is pictured in Figure 5.2. Note that in the domain is paired 
with two objects, 3 and 4, in the range. Hence, relation R is is not a function. 
As a second example, consider the relation T = {(5, 3), (6, 3), (7, 4)}, whose 
mapping diagram is pictured in Figure 5.3. In this example, each domain object 
is paired with exactly one range object: 5 only gets sent to 3, 6 only gets sent 
to 3, and 7 only gets sent to 4. Hence, the relation T is a function. The fact 
that the range object 3 is used twice does not matter. It's the fact that each 
domain object gets sent to exactly one range object that matters. 



You Try It! 



EXAMPLE 4. Consider the relation pictured in Figure 5.4. Is it a function? Consider the relation 

pictured below. Is it a 
y function? 



. 


, 


















E 










D 










B 












A 




C 
















1 

L i- 










5 

1 













/ 












































c 




























^ 








4 










































5 


















r 
















n 














R 














































r J 


■ 













-* X 



Figure 5.4: Is this relation a function? 



298 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Solution: The graph in Figure 5.4 consists of the points .4(1,1.), 5(2,2), 
C(3,2), £>(3,3), and £'(4,4). The domain is V = {1,2,3,4} and the range 
is 1Z = {1,2,3,4}. A mapping diagram (see Figure 5.5) will help us decide if 
the relation represented by the graph is a function. Put the domain on the left, 
the range on the right, then use arrows to indicate the ordered pairs. Let's call 
the relation /. 



/ 




Figure 5.5: A mapping diagram for the relation / depicted in Figure 5.4. 



Answer: Yes, the relation is 
a function. 



In Figure 5.5, note how the domain object 3 is "sent to" or paired with two 
range objects, 2 and 3. Hence, the relation / is not a function. 

□ 



You Try It! 



The following relation pairs 
people with their age. 
Determine if the relation is a 
function. 

S"={(Mary,23),(Joe, 18), 
(Alfonzo,20),(Zoe, 18)}, 
(Maria, 22), (Chris, 23)} 



EXAMPLE 5. The following relation pairs automobiles with their gas 
mileage. Determine if the relation is a function. 

T = {(Bentley Mulsanne, 18), (Kia Soul, 30), (Lamborghini Gallardo, 20), 
(Smart Fortwo,41), (Jaguar XF, 23)} 

Solution: In Figure 5.6, we create a mapping diagram indicating the relation 
between cars and their gas mileage. Note that each domain object on the left 
is paired with exactly one range object on the right. Hence, this relation is a 
function. 



Bentley Mulsanne 

Kia Soul 

Lamborghini Gallardo 

Smart Fortwo 

Jaguar XF 



-> 18 



-> 30 



-> 20 
-s-41 



-> 23 



Answer: Yes, the relation is 
a function. 



Figure 5.6: A mapping diagram for the relation T . 



□ 



5.1. FUNCTIONS 



299 



You Try It! 



EXAMPLE 6. The following relation pairs a particular bird with the state 
that has adopted that bird as its state bird. Determine if the relation is a 
function. 

R = { (Yellowhammer, Alabama), (Robin, Connecticut), 
(Nene, Hawaii), (Robin, Michigan)} 

Solution: In Figure 5.7, we create a mapping diagram indicating the relation 
between birds and their state adoptions. Note that the domain object "Robin" 
is paired with two range objects, "Connecticut" and "Michigan," hence this 
relation is not a function. 



R 



Yellowhammer 

Robin 

Nene 




-> Alabama 

> Connecticut 

> Hawaii 
Michigan 



The following relation pairs 
people with the types of cars 
they own. Determine if the 
relation is a function. 

S ={ (Bernard, station wagon), 
(Tina, truck), 
(Gilberto, sedan), 
(Kate, sport utility), 
(Bernard, sedan), 
(Kate, minivan)} 



Figure 5.7: A mapping diagram for the relation R. 



Answer: No, the relation is 
not a function. 



□ 



Mapping Diagram Notation 

The goal of this section is to introduce function notation. Let's begin with the 
mapping diagram in Figure 5.8. 



/ 



->2 
->6 



Figure 5.8: Mapping diagram . 



The mapping diagram in Figure 5.8 reveals the following facts: 

• / maps 1 to 2 or / : 1 — > 2. 

• / maps 2 to 4 or / : 2 -> 4. 



300 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



• f maps 3 to 6 or / : 3 — > 6. 

• / maps 4 to 8 or / : 4 -» 8. 



Note how the notation / : 4 — > 8 correlates nicely with the mapping diagram 
in Figure 5.8. The notation / : 4 — > 8 is read "/ maps 4 to 8" or "/ sends 4 to 
8." 

A closer look at the mapping diagram in Figure 5.8 reveals an interesting 
pattern. The "rule" seems to be that the relation / doubles each entry in its 
domain: twice 1 is 2, twice 2 is 4, twice 3 is 6, etc. It's possible to give a general 
description of this "rule" by writing: 



/ : x — > 2x 



(5.1) 



That is, / sends x to twice x, or equivalently, 2x. For example, we might ask 
"where does / send 15?" To answer this question, we would replace x with 15 
in the rule (5.1) to get 

/: 15 ->2(15), 



or equivalently, 



/: 15 -» 30. 



We could also ask "where does / send — 7?" To answer this question, we would 
replace x with —7 in the rule (5.1) to get 



or equivalently, 



/:-7->2(-7), 

/:-7->-14. 



You Try It! 



Given the rule 
/ : x — > 3x — 5, 
answer the question 
"where does / send —2?" 



Answer: / : — 2 



11 



EXAMPLE 7. Given the rule / : x — > 2x + 3, answer the question "where 
does / send 8?" 

Solution: To find where "/ sends 8," substitute 8 for x in the rule / : x — y 
2x + 3 to get 

/:8^2(8)+3, 



or equivalently, 



/:8->19. 



□ 



You Try It! 



Given the rule 
/ : x -» 2x 2 + 5x, 
answer the question 
"where does / send 3?" 



EXAMPLE 8. Given the rule / : x — > x/(x + 3), answer the question "where 
does / send —1?" 



5.1. FUNCTIONS 301 

Solution: To find where "/ sends — 1," substitute —1 for x in the rule 
/ : x — > x/(x + 3) to get 

/:-!■ "' 



-1 + 3 
or equivalently, 

1 2 



Answer: / : 3 — > 33 

D 



In Examples 7 and 8, note that each time you substitute a value for x in 
the given rule, you get a unique answer. This means that each object in the 
domain of / is sent to a unique object in the range of /, making the rules 
in Examples 7 and 8 functions. This leads us to an itemized description of a 
function. 

Rule of Three. A function consists of three parts: 

• a set of objects which mathematicians call the domain, 

• a second set of objects which mathematicians call the range. 



• and a rule that describes how to assign each object in the domain to 
exactly one object in the range. 



Function Notation 

Although the mapping diagram notation 

/ : x -» 3 - 4x 

is quite easy to understand, the standard function notation used is 

f(x) = 3 - Ax. 

With mapping diagram notation, if we want to answer the question "where 
does / send 12?" , we write: 



/ 


: x -> 3 - Ax 


/ 


12 -> 3-4(12) 


/ 


12 -> 3 - 48 


/ 


12 -» -45 



Hence, / : 12 — > —45; i.e., / sends 12 to —45. Function notation uses exactly 
the same concept; i.e, substitute 12 for x. 



302 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Tips for Using Function Notation. 

1 . Replace all occurrences of variables in the notation with open parentheses. 
Leave room between the parentheses to substitute the given value of the 
variable. 

2. Substitute the given values of variables in the open parentheses prepared 
in the first step. 

3. Evaluate the resulting expression according to the Rules Guiding Order 
of Operations. 



Given f(x) = 3 — 4x, to evaluate /(12), first restate the function notation, then 
replace each occurrence of the variable with open parentheses. 



f(x) = 3-Ax 
/( ) = 3-4( ) 



Original function notation. 

Replace each occurrence of x with 
open parentheses. 



Now substitute 12 for x in the open parentheses prepared in the last step. 



You Try It! 



4-5*<-3)-<-3) A 2 

10 



Figure 5.9: Calculator check. 



/(12) = 3-4(12) 



/(12) = 3-48 
/(12) = -45 

Hence, /(12) = -45; i.e., / sends 12 to -45 



Substitute 12 for x in the open 
parentheses positions. 

Multiply. 

Subtract. 



EXAMPLE 9. Given f{x) =A-5x-x 2 , evaluate /(-3). 

Solution: Start by replacing each occurrence of the variable x with open 
parentheses. 



/( ) 



bx — x 

5( )-( ) 2 



Original function notation. 

Replace each occurrence of x with 
open parentheses. 



Now substitute —3 for x in the open parentheses prepared in the last step. 



/(-3)=4-5(-3)-(-3) 2 



/(-3) 
/(-3) 
/(-3) 



4-5(-3)-9 

4+15-9 

10 



Substitute —3 for x in the open 
parentheses positions. 

Evaluate exponent: (—3) 

Multiply: -5(-3) = 15. 

Simplify. 



9. 



Thus, /(— 3) = 10. Check this on your calculator (see Figure 5.9). 



□ 



5.1. FUNCTIONS 



303 



The next example demonstrates one of the advantages of function notation. 
For example, it is easy to refer to the function in which you want to substitute 
the given x- value. 



You Try It! 



x and g(x) = x 2 — 9, find /(— 1) and Given f(x) = 3x 2 — 20 and 



EXAMPLE 10. Given f(x) = 
ff(-2). 

Solution: We're given two function definitions, / and g, but we're first asked 
to find /(— 1). This means that we must replace each occurrence of x with —1 
in the function f(x) = 5 — x. 



m 

/( ) 



X 

( ) 



Original function notation. 

Replace each occurrence of x with 
open parentheses. 



Now substitute —1 for x in the open parentheses prepared in the last step. 

/(-l) = 5-(-l) 



= 5 + 1 
= 6 



Substitute — 1 for x in the open 
parentheses position. 

Add the opposite. 

Simplify. 



Thus, /(— 1) = 6. We're next asked to find g{— 2). This means that we must 
replace each occurrence of x with —2 in the function g(x) = x 2 — 9. 



g(x) = x 2 - 
9( ) = ( ) 2 



9 



Original function notation. 

Replace each occurrence of x with 
open parentheses. 



Now substitute —2 for x in the open parentheses prepared in the last step. 



g(x) = 4x + 6/x, find /(-3) 
and g{2). 



5 (-2) = (-2) 2 


-9 


Substitute —2 for x in the open 
parentheses position. 






= 4-9 




Exponent first: (-2) 2 = 4. 






= -5 




Simplify. 






Thus, g{-2) = -5. 






Answer: 

/(-3) = 7 and g{2) = 


= 10 



□ 



Interchanging y and f{x) 

In most cases, y and f(x) are completely interchangeable. For example, com- 
pare and contrast the following two examples. 



304 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Sketch the graph of 
f{x) = -\x-2. 



Question: Given y 
y when x equals 5. 



3a; + 7, find 
Solution: Replace x with 5. 



y = 3x + 7 
y = 3(5) + 7 
y = 15 + 7 
V = 22 



Question: Given f(x) = 3x + 7, 
evaluate /(5). 

Solution: Replace x with 5. 

f(x) = 3ai + 7 
/(5) = 3(5) + 7 
/(5) = 15 + 7 
/(5) = 22 



In each case, the answer is 22. However, in the first case, the answer y = 22 
disguises the fact that an a;- value of 5 was used to arrive at the result. On 
the other hand, when we use function notation, the final answer /(5) = 22 
indicates that we used an a;- value of 5 to determine that the y- value is 22. This 
is another advantage of function notation. 

Let's look at one final application demonstrating that y and f(x) are inter- 
changeable. 



EXAMPLE 11. Sketch the graph of f(x) = 2x - 3. 

Solution: Because y and f(x) are interchangeable, the instruction is identical 
to "sketch the graph of y = 2x — 3." The graph is a line, with slope 2 and 
y-intercept at (0, —3). Plot the y-intercept at (0, —3), then move up 2 and right 
1 to create a line with slope 2 (see Figure 5.10). Note how we've labeled the 
graph with its equation using function notation. 



Answer: 











5 




























































































5 


















5 












[ [ 


'• - 


-Z 


) 












■~> 








































5 




3 




\ 


s 



f{x) = --x-2 











K ' 


i 








m 


















/ 






















/ 




















/ 






















/ 






























5 










1 








i 


























































(0 


— it 


) 
















^ 


■ 













2a: -3 



Figure 5.10: The graph of f(x) = 2x — 3 is a line. 



□ 



5.1. FUNCTIONS 



305 



**.!*.**. Exercises •** * ■** 



In Exercises 1-6, state the domain and range of the given relation. 



1. i? = {(7,4), (2,4), (4,2), (8,5)} 

2. 5 = {(6,4), (3,3), (2,5), (8,7)} 

3. T = {(7,2), (3,1), (9,4), (8,1)} 



4.7? = {(0,1), (8,2), (6,8), (9,3)} 

5. T = {(4,7), (4,8), (5,0), (0,7)} 

6. T = {(9,0), (3,6), (8,0), (3,8)} 



In Exercises 7-10, state the domain and range of the given relation. 



9. 











I 


/ 


























A 


































B 






























































5 


















! 






























C 










D 






































5 























r, 1 


/ 


















B 












































































A 




























5 


















! 
































c 














































5 










Z3 





c 



10. 



B 



c 



B 



D 



D 



In Exercises 11-18, determine whether the given relation is a function. 

11. R = {(-6, -4), (-4, -4), (1,-4)} 13. T = {(-1,-7), (2,-5), (4,-2)} 

12. T = {(-8, -3), (-4, -3), (2, -3)} 14. S = {(-6, -6), (-4, 0), (9, 1)} 



306 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



15. T= {(-9,1), (1,6), (1,8)} 

16. S = {(-7,0), (1,1), (1,2)} 



17. R = {(-7,-8), (-7,-6), (-5,0)} 

18. T = {(-8,-9), (-8, -4), (-5,9)} 



In Exercises 19-22, determine whether the given relation is a function. 
19. 21. 



.'/ 



. 


. 
































C 








B 










A 




















1 

L 1- 










5 
I 



20. 



1 


/ 








































































A 


B 


C 






1 

L 1- 










I 


' 



.'/ 



22. 































C 


















B 












A 








1 
L 1- 










5 

I 





/ 
















































C 














B 












A 








1 

L 1- 


■ 








5 
I 



23. Given f(x) 

24. Given f(x) 

25. Given f(x) 

26. Given f{x) 

27. Given f{x) 

28. Given f{x) 

29. Given f(x) 

30. Given f(x) 



|6x — 9|, evaluate /(8). 
: |8x — 3|, evaluate /(5). 
: -2x 2 + 8, evaluate /(3). 

3x 2 + x + 6, evaluate /(— 3). 

-3ie 2 +4ie+1, evaluate /(2). 

-3x 2 +4x-2, evaluate /(2). 
■ |5a; + 9|, evaluate /(-8). 
: \9x — 6|, evaluate /(4). 



\Jx — 6, evaluate /(42). 



Vx + 8, evaluate /(41) 



31. Given f(x 

32. Given f{x 

33. Given f(x) = y/x-7, evaluate /(88) 

34. Given f(x 

35. Given f(x 

36. Given f(x 

37. Given f(x 

38. Given f{x 



y/x + 9, evaluate /(16). 
— 4x + 6, evaluate /(8). 
— 9x + 2, evaluate /(— 6). 
— 6x + 7, evaluate /(8). 
—6x — 2, evaluate /(5). 



5.1. FUNCTIONS 



307 



39. Given f(x) = -2x 2 + 3x + 2 and g(x) = 
3x 2 + 5x — 5, evaluate /(3) and g(3). 

40. Given f(x) = 3a; 2 — 3x — 5 and g{x) = 
2x 2 — 5x — 8, evaluate /(— 2) and g(— 2). 

41. Given /(a;) = 6a; — 2 and g(a;) = —8a; + 9, 
evaluate /(— 7) and g(—7). 

42. Given /(a;) = 5a; — 3 and g(a:) = 9a; — 9, 
evaluate /(— 2) and g(— 2). 



43. Given /(x) = 4a; - 3 and g(a;) = -3a; + 8, 
evaluate /(— 3) and c/(— 3). 

44. Given f(x) = 8a; + 7 and g(x) = 2x - 7, 
evaluate /(— 9) and g(— 9). 

45. Given /(a;) = —2a; 2 + 5a; — 9 and g(x) = 
-2x 2 + 3a; - 4, evaluate /(-2) and g{-2). 

46. Given f(x) = -3a; 2 + 5a; - 2 and g(x) = 
3a; 2 — 4a; + 2, evaluate /(— 1) and g{— 1). 



ja- j*- &»■ Answers ••* *•* •** 



1. Domain = {2,4,7,8} and Range = 
{2,4,5} 

3. Domain = {3, 7, 8, 9} and Range = 
{1,2,4} 

5. Domain = {0, 4, 5} and Range = {0, 7, 8} 

7. Domain = { — 2,2} and Range = 
{-2,2,4} 

9. Domain = {—4,-1,1,2} and Range = 
{-2,2,4} 

11. Function 

13. Function 

15. Not a function 

17. Not a function 

19. Function 

21. Not a function 



23. 39 

25. -10 

27. -3 

29. 31 

31. 6 

33. 9 

35. -26 

37. -41 

39. /(3) = -7and ff (3) =37 

41. /(-7) = -44 and g(-7) = 65 

43. /(-3) = -15 and g(-3) = 17 

45. /(-2) = -27 and g{-2) = -18 



308 CHAPTER 5. POLYNOMIAL FUNCTIONS 

5.2 Polynomials 

We begin with the definition of a term. 

Term. A term is either a single number (called a constant term) or the product 
of a number and one or more variables. 

For example, each of the following is a term. 

-5 -3a; 2 12y 2 z 3 Ua 2 bc 3 

Note how the first term is a single number, while the remaining terms are 
products of a number and one or more variables. For example, — 3x 2 is the 
product of —3, x, and x. 

Coefficient. When a term is a product of a number and one or more variables, 
the number is called the coefficient of the term. In the case of a term that is 
a single number, the number itself is called the coefficient. 

Thus, for example, the coefficients of the terms 

-5 -3x 2 Yly 2 z 3 13a 2 6c 3 

are —5, —3, 12, and 13, respectively. 

Degree. The degree of a term is the sum of the exponents on each variable of 
the term. A constant term (single number with no variables) has degree zero. 

Thus, for example, the degrees of the terms 

-5 -3x 2 Yly 2 z 3 Ua 2 bc 3 

are 0, 2, 5, and 6, respectively. In the last example, note that 13a 2 6c 3 is 
equivalent to 13a 2 6 1 c 3 , so adding exponents, we get: 

Degree of 13a 2 6c 3 = Degree of I3a 2 b 1 c 3 
=2+1+3 
= 6 

Monomial. The words monomial and term are equivalent. 

Thus, 

-5 -3x 2 Yly 2 z 3 I3a 2 bc 3 

are monomials. 



5.2. POLYNOMIALS 309 



Binomial. A binomial is a mathematical expression containing exactly two 
terms, separated by plus or minus signs. 

For example, each of the mathematical expressions 

2x + 3y — 3a — 36 xy + 7 — 3x y + 5xy 

is a binomial. Each expression has exactly two terms. 

Trinomial. A trinomial is a mathematical expression containing exactly three 
terms, separated by plus or minus signs. 

For example, each of the mathematical expressions 

2x 2 + 3a; + 7 a 2 + 2ab + b 2 x A - 2x 2 y 2 + 3y 4 

is a trinomial. Each expression has exactly three terms. 

A bicycle has two wheels, a binomial has two terms. A tricycle has three 
wheels, a trinomial has three terms. But once we get past three terms, the 
assignment of special names ceases and we use the generic word polynomial, 
which means "many terms." 

Polynomial. A polynomial is a many-termed mathematical expression, with 
terms separated by plus or minus signs. The coefficients of a polynomial are 
the coefficients of its terms. 

Each of the previous expressions, 

12y 2 z 3 - 3a 2 - 3b 2 x 4 - 2x 2 y 2 + 3y A 

though assigned the particular names monomial, binomial, and trinomial, re- 
spectively, are also "many-termed" expressions and can also be called polyno- 
mials. However, because the word polynomial means "many terms," we can 
also use the word polynomial to describe mathematical expressions with more 
than three terms, such as: 

x 4 - 4x 3 y + 6x 2 y 2 - 4xy 3 + y A 

The coefficients of x A — 4x 3 y + 6x 2 y 2 — Axy 3 + y A are 1, —4, 6, —4, and 1. 

Ascending and Descending Powers 

When asked to simplify a polynomial expression, we should combine any like 
terms we find, and when possible, arrange the answer in ascending or descend- 
ing powers. 



310 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Simplify the following 
polynomial, and arrange 
your answer in ascending 
powers of x: 

3x 2 - 5a; 3 + 8x + 9a; 2 - 7x + 2x 3 



EXAMPLE 1. Simplify the following polynomial expression, arranging your 
answer in descending powers of x. Once you've completed that task, make a 
second arrangement, ordering your terms in ascending powers of x. 

2x 3 + 7x- 3x 2 + 11a; + 8a; 2 + 11 + 15a; 

Solution: In order to arrange our answer in descending powers of x, we want 
to place the term with the highest power of x first and the term with the lowest 
power of x last. We use the commutative and associative properties to change 
the order and regroup, then we combine like terms. 



2a; 3 + 7x - 3a; 2 + 11a; + 8a; 2 



11 + 15a; 



= 2a; 3 + (-3a; 2 + 8a; 2 ) + (7x + 11a: + 15a;) + 11 
= 2a; 3 + 5a; 2 + 33a; +11 

Note how the powers of x start at 3, then go down in order. 

To arrange our final answer in ascending powers of x, we put the lowest 
power of x first, then the highest power of x last, regrouping and combining 
like terms. 



Answer: x + 12a; 2 



33^ 



2a; 3 + 7x - 3x 2 + 11a; + 8a; 2 + 11 + 15a; 

= 11 + (7a; + 11a; + 15a;) + (-3a; 2 + 8a; 2 ) + 2a; 3 
= ll + 33x + 5x 2 + 2a; 3 

Note how we start with the constant term, then the powers of x increase in 
order. 

□ 



When we have a polynomial in a single variable, such as the polynomial 
in Example 1, arranging the terms in ascending or descending order is fairly 
straightforward. However, a polynomial in two or more variables is a bit more 
difficult, and sometimes impossible, to arrange in a decent order. 



You Try It! 



Simplify the following 
polynomial, and arrange 
your answer in descending 
powers of x: 
-ix 2 y 2 + 3xy 3 + 
6x 3 y — xy 3 + 2a; 2 y 2 



EXAMPLE 2. Simplify the following polynomial expression, then arrange 
your answer in descending powers of x. 



2xy 2 - 6x 2 y + y 3 



3xy 2 + 4x 2 y 



Solution: We'll again use the commutative and associative properties to 
change the order and regroup, putting the terms with the highest powers of x 



5.2. POLYNOMIALS 



311 



first, then follow with terms containing lower powers of x in order. 

x 3 + 2xy 2 - 6x 2 y + y 3 - 3xy 2 + 4x 2 y 

= x 3 + {-6x 2 y + 4x 2 y) + {2xy 2 - 3xy 2 ) + y 3 



x 3 - 2x 2 y 



xy^ + y J 



Note that this is a very natural order, the powers of x decrease while simulta- 
neously the powers of y increase. Answer: 

6x 3 y - 2x 2 y 2 + 2xy 3 

Not all examples will have nice ordering presented in Example 2, with the 
powers of one variable descending while the powers of the other variable simul- 
taneously ascends. Sometimes we have to make some very subjective choices 
on the ordering of terms. 



□ 



EXAMPLE 3. Simplify the following polynomial expression, then arrange 
your answer in some sort of reasonable order. 



a 3 b 3 + 2a 2 b - 3a 2 b 3 + Aa 3 b 3 + 5a 4 + 3a A b + 6° 

Solution: Let's try to arrange the terms so that the powers of a descend. 
Again, we use the commutative and associative properties to change the order 
and regroup. 



You Try It! 



Simplify the following 
polynomial, and arrange 
your answer in ascending 
powers of b: 
5a 3 b 2 + 4ab 3 - 2a 2 b+ 
3a 3 b 2 - ab 3 



3^3 



(!%■ 



2a 2 b- 



3a 2 b 3 



Aa 3 b 3 + 5a 4 + 3a 2 b + b 5 



= 5a 4 + (a 3 b 3 + 4a 3 b 3 ) + (2a 2 b + 3a 2 b) - 3a 2 b 3 + b b 
= 5a 4 + 5a 3 6 3 + 5a 2 b - 3a 2 b 3 + b 5 

Note that in our final arrangement, the powers of a descend, but the powers of 
b bounce up and down, but at least we have the powers of a descending. That 
should help us spot if we've missed a term while simplifying the given problem. 



Answer: 
-2a 2 b + 8a 3 b 2 



3ab 3 



□ 



The Degree of a Polynomial 

To find the degree of a polynomial, locate the term of the polynomial having 
the highest degree. 



The degree of a polynomial. The 

the term having the highest degree. 



of a polynomial is the degree of 



312 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



What is the degree of the 

polynomial 

2a; 3 + 8x 2 + 3a; 4 + 2x + 10? 



Answer: 4 



You Try It! 



What is the degree of the 

polynomial 

x 2 y 4 — 6x 2 y 2 + 5x 2 y 5 — 2xy? 



Answer: 7 



Finding the degree of a polynomial of a single variable is pretty easy. 

EXAMPLE 4. What is the degree of the polynomial a: 3 — 4a; 2 + 5 — 6x + 2x 7 ? 
Solution: First, let's arrange the polynomial in descending powers of x. 



2x' 



4cT - 6x + 5 



Arranging the polynomial in descending powers of x makes it easier to see 
that the term of the polynomial with the highest degree is 2a; 7 . Therefore, the 
degree of the polynomial is 7. 

□ 

Finding the degree of a polynomial of more than one variable is a little bit 
trickier. 



EXAMPLE 5. What is the degree of the polynomial x — 2x y 



3„,7 



y 



5? 



Solution: Note that the polynomial is already arranged in descending powers 
of a;, an arrangement that is probably as good as we are going to get. In the 
following table, we list the degree of each term. Remember, the degree of any 
term is found by summing the exponents on its variables. 

Term Degree 



x 4 


4 


-2x i y 7 


10 


y 5 


5 



Hence, the term with the highest degree is —2x 3 y 7 , making 10 the degree of 
the polynomial. 

□ 



Polynomial Functions 

First we define what we mean by a polynomial function. 



Polynomial function. A polynomial function is a function defined by a rule 
that assigns to each domain object a range object defined by a polynomial 
expression. 



5.2. POLYNOMIALS 



313 



Advanced courses, such as multivariate calculus, frequently use polynomial 
functions of more than one variable such as f(x, y) = x 2 + y 2 . However, in this 
course, our focus will be on polynomial functions of a single variable, such as 
p(x) = 3 - ix - 9x 2 and q(x) = x 3 - 9x 2 + 11. 



You Try It! 



EXAMPLE 6. Given the polynomial function p(x) 
p(-3). 



Solution: To evaluate p(— 3), first restate the function definition, then replace 
each occurrence of the variable x with open parentheses. 



8a; — 11, evaluate Given the polynomial 
function 

p(x) = -3x 2 + 7x + A, 
evaluate p(2). 



p(x) = x — 8x 
p( ) = ( ) 3 " 



11 Original function definition. 

( ) — 1 1 Replace each occurrence of x with 

open parentheses. 



Next, substitute —3 for x in the open parentheses prepared in the last step. 



p(-3) = (-3) s 



p(-3) = 
p(-3) = 
p(-3) = 

Hence, p(— 3) 
Figure 5.11). 



-27-8(-3)- 11 

-27+24- 11 
-14 



11 Substitute —3 for x in the open 

parentheses positions. 

Exponent first: (-3) 3 = -27. 

Multiply: -8(-3) = 24. 

Add. 



( -3)^3-8* <-3)-ll 
-14 



Figure 5.11: 
check. 



Calculator 



-14. You can easily check this result on your calculator (see 



Answer: 6 



The Graph of a Polynomial Function 

One of the most important polynomial functions in all of mathematics and 
science is the polynomial having degree two. 



Quadratic polynomial. The second degree polynomial having the form 

p(x) = ax 2 + bx + c 

is called a quadratic polynomial. The graph of this polynomial is called a 
parabola. 



□ 



The parabola is approximately U-shaped. Some open upwards, some open 
downwards, depending on the sign of the leading term. In Figure 5.12, the 
leading term of the parabola p(x) = 2x 2 — 8x + 6 has positive two as its 



314 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



-5-* 

Figure 5.12: The graph of p(x) 
2x 2 — 8x + 6 opens up. 



H 5 "' 




zx 


«_ r\ 


sir T 


V-4- t 


J t 


=fc=fe= 



Figure 5.13: The graph of p(x) 
—2x 2 — 8a; — 6 opens down. 



coefficient, so it opens upward. In Figure 5.13, the leading term of the parabola 
p(x) = — 2x 2 — 8x — 6 has negative two as its coefficient, so it opens downward. 



Thes 


tgn of the leading term of 


p(x) = ax 2 


+ bx + c 


determines 


whether the 


parabola 


opens up or down. 










• If a 


> 0, the parabola opens 


upward. 








• If a 


< 0, the parabola opens 


downward. 









The turning point of a parabola has a special name. 

The vertex of a parabola. The graph of the second degree polynomial 
p(x) = ax 2 + bx + c has a single turning point, called the vertex of the parabola. 



You Try It! 



Use your graphing calculator 
to sketch the graph of the 
quadratic polynomial 
p(x) = 2x 2 - 5x - 4. 



EXAMPLE 7. Use your graphing calculator to sketch the graph of the 
quadratic polynomial p(x) = —3a; 2 + 12a; + 25. 

Solution: The degree of the polynomial p(x) = —3x 2 + 12a; + 25 is two, so 
it is a quadratic polynomial and its graph is a parabola. Moreover, its leading 
term has negative three as its coefficient, so we know that the parabola opens 
downward. Enter y = -3a; 2 + 12a; + 25 as Yl=-3*XA2+12*X+25 in the 
Y= menu (see the first image in Figure 5.14), then select 6:ZStandard from 
the ZOOM menu to produce the third image in Figure 5.14. 

Note that the graph in Figure 5.14 appears to have the U-shape of a parabola 
that opens downwards. Its vertex (turning point) is not visible, but one would 



5.2. POLYNOMIALS 



315 



Plotl PlotE Plots 
WiB-3*X A 2+12*X+ 
25 

sVs = 



MEMORV 
ox 
Zoom In 
Zoom Out 
ZDecinal 
ZS=iuare 
ZStandard 



7-i-ZTrig 



Figure 5.14: Sketching the graph of p(x) = —3x 2 + 12a; + 25. 



surmise that it lies off the top of the screen. We need to adjust the WINDOW 
parameters so that the vertex of the parabola is visible in the viewing screen. 
After some experimentation, we settle on the parameters shown in the first 
image in Figure 5.15, then push the GRAPH button to produce the second 
image in Figure 5.15. 



WINDOW 
Xnin=-10 
Xmax=10 
Xscl=l 
Ymin=-50 
Vmax=50 
Yscl=10 

4rXres=i 




Figure 5.15: Adjust the WINDOW parameters so that the vertex is visible in 
the viewing screen. 

In reporting your result on your homework, follow the Calculator Submis- 
sion Guidelines from Chapter 3, Section2. 



!J 



1. Draw axes with a ruler. 

2. Label the horizontal axis x 
and the vertical axis y. 

3. Indicate the WINDOW 
parameters Xmin, Xmax, 
Ymin, and Ymax at the 

end of each axis. 

4. Freehand the curve and label 
it with its equation. 





50 


p{x) = 


-3x 2 


10 


'-50" 




10 



12a: + 25 



Answer: 




□ 



316 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Use your graphing calculator 
to sketch the graph of the 
quadratic polynomial 
p(x) = x 3 - 14a; 2 + 20a; + 60. 
Set your window parameters 
as follows: Xmin = —10, 
Xmax = 20, Xscl = 1, 
Ymin = -200, 
Ymax = 200, and 
Yscl = 20. 



Answer 




■ ■ / 


^ / 


/I 


: V 



When the degree of the polynomial is larger than two, the number of turning 
points of the graph might increase. This makes for some very interesting curves. 
In more advanced courses, such as intermediate and college algebra, you will be 
introduced to a variety of techniques that will help you determine appropriate 
viewing windows for the graphs of these higher degree polynomials. However, 
in this introductory section, we will assist you by suggesting a good viewing 
window for each polynomial, one that will allow you to see all of the turning 
points of the graph of the polynomial. 



EXAMPLE 8. Use your graphing calculator to sketch the graph of the 
polynomial function p(x) = x A — 37a; 2 + 24a; + 180. Set your window parameters 
as follows: Xmin = —10, Xmax = 10, Xscl = 1, Ymin = —1000, Ymax = 
1000, and Yscl = 100. 

Solution: Enter the polynomial function in Yl of the Y= menu, then enter 
the suggested window parameters in the WINDOW menu (see Figure 5.16). 



Plotl PlotE Plots 


WINDOW 


WiBX-M-37*>T-2+2 


Xnin=-10 


4*X+180 


Xnax=l@ 


Wi=l 


Xscl=l 


^Vs= 


Ymin=-1000 


Wh = 


Vnax=1000 


^Ve= 


Yscl=100 


Ws = 


l¥.res=W 



Figure 5.16: Enter the polynomial and adjust the WINDOW parameters. 

Push the GRAPH button on the top row of your calculator to produce the 
graph of the polynomial function shown in Figure 5.17. 




Figure 5.17: The graph of p(x) = x 4 - 37a; 2 + 24a; + 180. 



Sweet-looking curve! 



□ 



5.2. POLYNOMIALS 317 



**.**.**. Exercises -*s -« ■*? 

In Exercises 1-6, state the coefficient and the degree of each of the following terms. 

1. 3v 5 u 6 4. -5c 3 

2. -36 5 ;z 8 5. 2u 7 x i d 5 

3. -5v 6 6. to 4 c 5 u 7 

In Exercises 7-16, state whether each of the following expressions is a monomial, binomial, or trinomial. 

7. -76 9 c 3 12. 8w 4 + buv + 3v 4 

8. 76 6 c 2 13. 5s 2 + 9£ 7 

9. Au + 7v 14. -8a; 6 - 6y 7 

10. -36 + 5c 15. 2u 3 - buv - 4v 4 

11. 36 4 - 96c + 9c 2 16. 6y 3 - Ayz + 7z 3 

In Exercises 17-20, sort each of the given polynomials in descending powers of x. 

17. -2a; 7 - 9a; 13 - 6a; 12 - 7a; 17 19. 8a; 6 + 2a; 15 - 3a; 11 - 2a; 2 

18. 2a; 4 - 8a; 19 + 3a; 10 - 4a; 2 20. 2a; 6 - 6a; 7 - 7a; 15 - 9a; 18 



In Exercises 21-24, sort each of the given polynomials in ascending powers of x. 

21. 7a; 17 + 3a; 4 - 2a; 12 + 8a; 14 23. 2a; 13 + 3a; 18 + 8a; 7 + 5a; 4 

22. 6a; 18 - 6a; 4 - 2a; 19 - 7a; 14 24. -6x 18 - 8a; 11 - 9a; 15 + 5a; 12 



In Exercises 25-32, simplify the given polynomial, combining like terms, then arranging your answer in 
descending powers of x. 

25. -5a; + 3 - 6x 3 + 5a; 2 - 9x + 3 - 3a; 2 + 6a; 3 29. a; 2 + 9a; - 3 + 7a; 2 - 3a; - 8 

26. -2a; 3 + 8a; - a; 2 + 5 + 7 + 6a; 2 + 4a; 3 - 9a; 30. -4a; 2 - 6a; + 3 - 3a; 2 + 3a; - 6 

27. 4a; 3 + 6a; 2 - 8a; + 1 + 8a; 3 - 7a; 2 + 5a; - 8 31. 8a; + 7 + 2a; 2 - 8a; - 3a; 3 - a; 2 

28. -8a; 3 - 2a: 2 - 7a; - 3 + 7a; 3 - 9a; 2 - 8a; + 9 32. -a; 2 + 8 - 7a; + 8a; - 5a; 2 + 4a; 3 



318 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



In Exercises 33-44, simplify the given polynomial, combining like terms, then arranging your answer 
in a reasonable order, perhaps in descending powers of either variable. Note: Answers may vary, 
depending on which variable you choose to dictate the order. 



33. -8a; 2 - 4xz - 2z 2 - 3a; 2 - 8xz + 2z 2 

34. -5a; 2 + 9xz - Az 2 - 6a; 2 - Ixz + 7z 2 

35. -6u 3 + Auv 2 - 2v 3 - u 3 + 6u 2 v — buv 2 



36. 7a 3 



6a 2 



bab 2 + 4a 3 



6a 2 



6b 3 



37. -46 2 c - 36c 2 - 5c 3 + 96 3 - 36 2 c + 56c 2 

38. 46 3 - 66 2 c + 96c 2 - 96 3 - 86c 2 + 3c 3 



39. -8y 2 + 6yz - 7z 2 - 2y 2 - 3yz - 9z 2 

40. 8x 2 + xy + 3y 2 - x 2 + 7xy + y 2 

41. 76 2 c + 86c 2 - 6c 3 - 46 3 + 96c 2 - 6c 3 

42. 7x 3 - 9x 2 y + 3y 3 + 7x 3 + 3xy 2 - 7y 3 

43. 9a 2 + ac- 9c 2 - 5a 2 - 2ac + 2c 2 

44. 7m 2 + 3uv - 6v 2 - 6u 2 + 7uv + 6v 2 



In Exercises 45-50, state the degree of the given polynomial. 



45. 3x 15 + 4 + 8x 3 - 8x 19 

46. -4a; 6 - 7ir 16 - 5 + 3:r 18 

47. 7a; 10 - 3a; 18 + 9a; 4 - 6 



48. 3a; 16 

49. -2 - x' - 5a; D + x^ 

50. x 11 + 7a; 16 + 8 - 7a; 10 



8x 5 

n7 



8 + 7 

10 



51. Given f(x) = 5x 3 + Ax 2 — 6, evaluate 
/(-I)- 

52. Given f(x) = -3a; 3 + 3a; 2 - 9, evaluate 
/(-I)- 

53. Given f(x) = 5a; 4 — 4a; — 6, evaluate /(— 2). 

54. Given f{x) = -2x 4 -4x-9, evaluate /(2). 

55. Given f(x) = 3a; + 5a; 3 — 9, evaluate 

/(-2). 



56. Given f(x) = —3a; 4 + 2a; 3 — 6, evaluate 
/(-I)- 

57. Given f(x) = 3a; 4 — 5a; 2 + 8, evaluate 
/("!)■ 

58. Given f(x) = —4a; 4 — 5a; 2 — 3, evaluate 
/(3). 

59. Given f(x) = -2a; 3 +4a;-9, evaluate /(2). 

60. Given f(x) = 4a; 3 +3a;+7, evaluate f{-2). 



In Exercises 61-64, use your graphing calculator to sketch the the given quadratic polynomial. In each 
case the graph is a parabola, so adjust the WINDOW parameters until the vertex is visible in the 
viewing window, then follow the Calculator Submission Guidelines when reporting your solution on 
your homework. 



61. p(x) = -2x 2 + 8x + 32 

62. p(x) = 2x 2 + &x- 18 



63. p(x) = 3a; 2 - 8a; - 35 

64. p(x) = -Ax 2 - 9a; + 50 



5.2. POLYNOMIALS 



319 



In Exercises 65-68, use your graphing calculator to sketch the polynomial using the given WINDOW 
parameters. Follow the Calculator Submission Guidelines when reporting your solution on your home- 
work. 



65. p(x) 



x 

Xmin 
Ymin 



66. p{x) = — x 3 

Xmin = 
Ymin = 



Ax 1 



-10 

-50 

4a; 2 



11a; + 30 

Xmax = 
Ymax 



-10 



-10 

-150 



50 



27a: - 90 

Xmax= -10 
Ymax = 50 



67. p(x) 



x 



10a; 3 - Ax 1 + 250a; - 525 



Xmin 
Ymin 



-10 
-1000 



Xmax= -10 
Ymax = 500 



68. p(x) 



-x 



2x 6 + 35a; 2 - 36a; - 180 



Xmin 
Ymin 



-10 

-50 



Xmax 
Ymax 



10 



50 



j*- j*- ?*- Answers •** ■*$ ■** 



1. Coefficient = 3, Degree =11 
3. Coefficient = — 5, Degree = 6 
5. Coefficient = 2, Degree = 16 
7. Monomial 
9. Binomial 
11. Trinomial 
13. Binomial 
15. Trinomial 

17. -7a; 17 - 9a; 13 - 6a; 12 - 2a; 7 
19. 2a; 15 - 3a; 11 + 8a; 6 - 2a; 2 
21. 3a; 4 - 2a; 12 + 8a; 14 + 7a; 17 
23. 5a; 4 + 8a; 7 + 2a; 13 + 3a; 18 
25. 2a; 2 - 14a; + 6 
27. 12a; 3 - a; 2 - 3a; - 7 



29. 8a; 2 + 6a;- 11 

31. -3ar 3 +a; 2 + 7 

33. -11a; 2 - 12a;z 

35. -7u 3 + 6u 2 v - uv 2 - 2v 3 

37. 96 3 - 76 2 c + 26c 2 - 5c 3 

39. -Wy 2 + 3yz- I6z 2 

41. -46 3 + 76 2 c + 176c 2 - 12c 3 

43. 4a 2 -ac- 7c 2 

45. 19 

47. 18 

49. 10 

51. -7 

53. 82 

55. -1 

57. 6 



320 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



63. 



p(x) = -2x 2 + 8x + 32 




65. 




p(x) = -3x 2 - 8a; - 35 



67. 



p(x) = x 3 - 4x 2 - Ux + 40 

y 




p(x) = x 4 - 10x 3 - 4x 2 + 250a: - 525 
V 




1000 



5.3. APPLICATIONS OF POLYNOMIALS 



321 



5.3 Applications of Polynomials 

In this section we investigate real-world applications of polynomial functions. 



EXAMPLE 1. The average price of a gallon of gas at the beginning of each 
month for the period starting in November 2010 and ending in May 2011 are 
given in the margin. The data is plotted in Figure 5.18 and fitted with the 
following third degree polynomial, where t is the number of months that have 
passed since October of 2010. 

p(t) = -0.0080556i 3 + 0.11881£ 2 - 0.30671* + 3.36 (5.2) 

Use the graph and then the polynomial to estimate the price of a gallon of gas 
in California in February 2011. 



You Try It! 



p(t) 


































4.50 


. 






















































4.40 






































































4.30 






































































4.20 






































































4.10 






































































4 






































































3.90 






















































i 


» / 














3.80 






































































3.70 






































































3.60 






































































3.50 






































































3.40 






































































3.30 


V 


































\ 


































3.20 
3.10 




^ 


































i 


F — 
































































3 
( 




































) 


1 


2 


3 


4 


5 


6 


7 


8 





ci 


N 


)V 


D 


ec 


L 


in 


F( 


;b 


M 


ar 


A 


jr 


M 


ay 


J i 


in 



Month 


Price 


Nov. 


3.14 


Dec. 


3.21 


Jan 


3.31 


Mar. 


3.87 


Apr. 


4.06 


May 


4.26 



Figure 5.18: Fitting gas price versus month with a cubic polynomial. 



Solution: Locate February (t = 4) on the horizontal axis. From there, draw 
a vertical arrow up to the graph, and from that point of intersection, a second 
horizontal arrow over to the vertical axis (see Figure 5.19). It would appear 
that the price per gallon in February was approximately $3.51. 



322 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



p(t) 


































4.50 


. 






















































4.40 






































































4.30 






































































4.20 






































































4.10 






































































4 






































































3.90 






















































<> / 














3.80 






































































3.70 






































































3.60 
3.50 
3.40 










































































































































3.30 






































































3.20 






































































3.10 




i 


> — 
































































3 
( 




































) 


1 


2 


3 


4 


5 


6 


7 


8 





ct 


N 


)V 


D 


ec 


Jc 


in 


F< 


sb 


M 


ar 


Apr 


May 


Jun 



Figure 5.19: Approximating price of gas during February. 

Next, we'll use the fitted third degree polynomial to approximate the price 
per gallon for the month of February, 2011. Start with the function defined by 
equation 5.2 and substitute 4 for t. 

p(t) = -0.0080556t 3 + 0.11881* 2 - 0.30671* + 3.36 

p(4) = -0.0080556(4) 3 + 0.11881(4) 2 - 0.30671(4) + 3.36 

Use the calculator to evaluate p(4) (see Figure 5.20). Rounding to the nearest 



-0.0080556*4^3+0 
.11331*4^2-0.306 
71*4+3.36 

3.5185616 



Figure 5.20: Evaluating p (4). 



penny, the price in February was $3.52 per gallon. 



□ 



5.3. APPLICATIONS OF POLYNOMIALS 



323 



EXAMPLE 2. If a projectile is fired into the air, its height above ground at 
any time is given by the formula 



where 



1 2 

y = yo + v t - -gt 



y = height above ground at time t, 
2/o = initial height above ground at time t = 0, 
vq = initial velocity at time t = 0, 
g = acceleration due to gravity, 
t = time passed since projectile's firing. 



(5.3) 



You Try It! 



If a projectile is launched 
with an initial velocity of 60 
meters per second from a 
rooftop 12 meters above 
ground level, at what time 
will the projectile first reach 
a height of 150 meters? 



If a projectile is launched with an initial velocity of 100 meters per second 
(100 m/s) from a rooftop 8 meters (8 m) above ground level, at what time will 
the projectile first reach a height of 400 meters (400m)? Note: Near the earth's 
surface, the acceleration due to gravity is approximately 9.8 meters per second 
per second (9.8 (m/s)/s or 9.8 m/s 2 ). 



Solution: We're given the initial height is j/o = 8m 
vq = 100 m/s, and the acceleration due to gravity is g 



the initial velocity is 
9.8 m/s 2 . Substitute 
these values in equation 5.3, then simplify to produce the following result: 



y = yo + v t 



1 2 

-gt 



y = 8 + 100i - -(9-8)t 2 
y = 8 + WOt - 4.9i 2 



Enter y 



lOOt - AM 2 as Y1=8+100*X-4.9*XA2 in the Y= menu (see 



the first image in Figure 5.21). After some experimentation, we settled on the 
WINDOW parameters shown in the second image in Figure 5.21. Push the 
GRAPH button to produce the graph of y = 8 + lOOt — 4.9£ 2 shown in the 
third image Figure 5.21. 



In this example, the horizon- 
tal axis is actually the t-axis. 
So when we set Xmin and 
Xmax, we're actually setting 
bounds on the t-axis. 



Plotl PlotE Plots 
WiB8+100*X-4.9* 

nVs= 



WINDOW 
Xnin=0 
Xnax=25 
Xscl=5 
Ymin=-100 
Ymax=600 
Yscl=50 

4,Xres=l 




Figure 5.21: Sketching the graph of y = 8 + WOt - 4.9t 2 . 



324 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



To find when the projectile reaches a height of 400 meters (400 m), substi- 
tute 400 for y to obtain: 



400 = 8 + 100* - 4.9r 



(5.4) 



Enter the left-hand side of Equation 5.4 into Y2 in the Y= menu, as shown in 
the first image in Figure 5.22. Push the GRAPH button to produce the result 
shown in the second image in Figure 5.22. Note that there are two points of 
intersection, which makes sense as the projectile hits 400 meters on the way 
up and 400 meters on the way down. 



Plotl PlotE Plots 
WiB8+100*X-4.9* 

vViB400 
sVs=i 

Wh = 





Intersection 
K=£.292£3£9 Y=400 



Figure 5.22: Determining when the object first reaches 400 meters. 



The parabola shown in 
Figure 5.23 is not the actual 
flight path of the projectile. 
The graph only predicts the 
height of the projectile as a 
function of time. 



To find the first point of intersection, select 5:intersect from the CALC 
menu. Press ENTER in response to "First curve," then press ENTER again 
in response to "Second curve." For your guess, use the arrow keys to move 
the cursor closer to the first point of intersection than the second. At this 
point, press ENTER in response to "Guess." The result is shown in the third 
image in Figure 5.22. The projectile first reaches a height of 400 meters at 
approximately 5.2925359 seconds after launch. 




-100 



5.2925359 



Figure 5.23: Reporting your graphical solution on your homework. 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 



5.3. APPLICATIONS OF POLYNOMIALS 



325 



• Label the horizontal and vertical axes with t and y, respectively (see 
Figure 5.23). Include the units (seconds (s) and meters (m)). 

• Place your WINDOW parameters at the end of each axis (see Figure 5.23) . 
Include the units (seconds (s) and meters (m)). 

• Label each graph with its equation (see Figure 5.23). 

• Draw a dashed vertical line through the first point of intersection. Shade 
and label the point (with its t- value) where the dashed vertical line crosses 
the i-axis. This is the first solution of the equation 400 = 8 + lOOt — AM 2 
(see Figure 5.23). 

Rounding to the nearest tenth of a second, it takes the projectile approximately 
t ~ 5.3 seconds to first reach a height of 400 meters. 



The phrase "shade and label 
the point" means fill in the 
point on the i-axis, then write 
the i-value of the point just 
below the shaded point. 

Answer: 

pa 3.0693987 seconds 



□ 



Zeros and x-intercepts of a Function 

Recall that f(x) and y are interchangeable. Therefore, if we are asked to find 
where a function is equal to zero, then we need to find the points on the graph 
of the function that have a y- value equal to zero (see Figure 5.24). 











. 


, 








f 




























































































(-3 


0) 






(" 


1,0 


) 




|(3, 


0) 




r 
-5 














/ 




i 


















/ 


















\ 




/ 


















\ 


/ 


/ 
















-5- 















Figure 5.24: Locating the zeros of a function. 



Zeros and x-intercepts. The points where the graph of / crosses the x-axis 
are called the x-intercepts of the graph of /. The x-value of each x-intercept 
is called a zero of the function /. 



326 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



The graph of / crosses the x-axis in Figure 5.24 at (—3,0), (—1,0), and 
(3, 0). Therefore: 

• The x-intercepts of / are: (—3,0), (—1,0), and (3,0) 

• The zeros of / are: —3, —1, and 3 

Key idea. A function is zero where its graph crosses the x-axis. 



You Try It! 



Find the zero(s) of the 
function f(x) = 2.6x — 9.62. 



EXAMPLE 3. Find the zero(s) of the function f(x) = 1.5a; + 5.25. 

Algebraic solution: Remember, /(x) = 1.5a; + 5.25 and y = 1.5a; + 5.25 are 
equivalent. We're looking for the value of x that makes y = or f(x) = 0. So, 
we'll start with fix) = 0, then replace /(x) with 1.5a; + 5.25. 

We want the value of x that makes 
the function equal to zero. 

Replace f{x) with 1.5a; + 5.25. 





/(*) = 







1.5a; + 5.25 = 





Now 


we solve for x. 






1.5a; = 


-5.25 




x = 


-5.25 

1.5 




x = 


-3.5 



Subtract 5.25 from both sides. 
Divide both sides by 1.5. 
Divide: -5.25/1.5 = -3.5 

Check. Substitute —3.5 for x in the function f(x) = 1.5a; + 5.25. 



/(x) = 1.5a; + 5.25 
/(-3.5) = 1.5(-3.5) + 5.25 
/(-3.5) = -5.25 + 5.25 
/(-3.5) = 



The original function. 
Substitute —3.5 for x. 
Multiply: 1.5(-3.5) = -5.25. 
Add. 



Note that —3.5, when substituted for x, makes the function f(x) = 1.5a; + 5.25 
equal to zero. This is why —3.5 is called a zero of the function. 

Graphing calculator solution. We should be able to find the zero by sketch- 
ing the graph of / and noting where it crosses the x-axis. Start by loading the 
function f(x) = 1.5x + 5.25 into Yl in the Y= menu (see the first image in 
Figure 5.25). 

Select 6:ZStandard from the ZOOM menu to produce the graph of / 
(see the second image in Figure 5.25). Press 2ND CALC to open the the 
CALCULATE menu (see the third image in Figure 5.25). To find the zero of 
the function /: 



5.3. APPLICATIONS OF POLYNOMIALS 



327 



Plotl PlotE Plots 
WiB1.5*X+5.25 
-■■Vi = 




gilEfiilEiU? 



1 : yalue 
zero 
: riininuri 
4: maximun 
5: inter-sect 
fcidy/dx 
7:^fCx>dx 



Figure 5.25: Finding the zero of f(x) = 1.5a; + 5.25. 



1. Select 2:zero from the CALCULATE menu. The calculator responds 
by asking for a "Left Bound?" (see the first image in Figure 5.26). Use 
the left arrow button to move the cursor so that it lies to the left of the 
IE-intercept of / and press ENTER. 

2. The calculator responds by asking for a "Right Bound?" (see the second 
image in Figure 5.26). Use the right arrow button to move the cursor so 
that it lies to the right of the x-intercept of / and press ENTER. 

3. The calculator responds by asking for a "Guess?" (see the third image in 
Figure 5.26). As long as your cursor lies between the left- and right-bound 
marks at the top of the screen (see the third image in Figure 5.26), you 
have a valid guess. Since the cursor already lies between the left- and 
right-bounds, simply press ENTER to use the current position of the 
cursor as your guess. 



Y1=i.E*K+E.£! 




Left Bound? 
K=0 


Y=£.2£ 



Y1=i.E*K+E.£E 




/ 

RiSht Bound? 

K=-4.2££319 


Y=-i.i32979 



Y1=i.E*K+E.£E 

a 




/ 

Gut::? 
K=-2.££319i 


Y=1.H20212H 



Figure 5.26: Using 2:zero from the CALCULATE menu. 

The calculator responds by approximating the zero of the function as shown in 
Figure 5.27. 







/ 

StKO 
K=-3.£ 


Y=0 



Figure 5.27: —3.5 is a zero of /. 

Note that the approximation found using the calculator agrees nicely with the 
zero found using the algebraic technique. 



Answer: 3.7 



□ 



328 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



If a projectile is launched 
with an initial velocity of 60 
meters per second from a 
rooftop 12 meters above 
ground level, at what time 
will the projectile return to 
ground level? 



In this example, the 
horizontal axis is actually 
the i-axis. So when we set 
Xmin and Xmax, we're 
actually setting bounds on 
the t-axis. 



EXAMPLE 4. How long will it take the projectile in Example 2 to return 
to ground level? 

Solution: In Example 2, the height of the projectile above the ground as a 
function of time is given by the equation 

y = 8 + 100t- AM 2 . 

When the projectile returns to the ground, its height above ground will be 
zero meters. To find the time that this happens, substitute y = in the last 
equation and solve for t. 







100t-4.9r 



Enter the equation y = 8 + 100£ — 4.9£ 2 into Yl in the Y= menu of your 
calculator (see the first image in Figure 5.28), then set the WINDOW param- 
eters shown in the second image in Figure 5.28. Push the GRAPH button to 
produce the graph of the function shown in the third image in Figure 5.28. 



Plotl PlotE Plots 
WiB8+100*X-4.9* 

Wh = 



WINDOW 
Xnin=0 
Xmax=25 
Xscl=5 
Ymin=-100 
Ymax=600 
Yscl=50 

l¥.res=W 




Figure 5.28: Sketching the graph of y 



lOOi - 4.9^ 



To find the time when the projectile returns to ground level, we need to find 
where the graph of y = 8 + lOOt — A.9t 2 crosses the horizontal axis (in this case 
the i-axis) . Select 2:zero from the CALC menu. Use the arrow keys to move 
the cursor slightly to the left of the ^-intercept, then press ENTER in response 
to "Left bound." Move your cursor slightly to the right of the ^-intercept, then 
press ENTER in response to "Right bound." Leave your cursor where it is and 
press ENTER in response to "Guess." The result is shown in Figure 5.29. 




Figure 5.29: Finding the time when the projectile hits the ground. 



5.3. APPLICATIONS OF POLYNOMIALS 



329 



• Label the horizontal and vertical axes with t and y, respectively (see 
Figure 5.30). Include the units (seconds (s) and meters (m)). 

• Place your WINDOW parameters at the end of each axis (see Figure 5.30). 

• Label the graph with its equation (see Figure 5.30). 

• Draw a dashed vertical line through the ^-intercept. Shade and label the 
i-value of the point where the dashed vertical line crosses the i-axis. This 
is the solution of the equation = 8 + lOOi — 4.9f 2 (see Figure 5.30). 



y(m) 
600 1 



-100 




20.487852 



Figure 5.30: Reporting your graphical solution on your homework. 



Rounding to the nearest tenth of a second, it takes the projectile approximately 

t rs 20.5 seconds to hit the ground. Answer: 

« 12.441734 seconds 



□ 



330 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



;».;-». t± 



Exercises 



•*;■*; -*i 



1. A firm collects data on the amount it spends on advertising and the resulting revenue collected 
by the firm. Both pieces of data are in thousands of dollars. 



x (advertising costs) 





5 


15 


20 


25 


30 


R (revenue) 


6347 


6524 


7591 


8251 


7623 


7478 



The data is plotted then fitted with the following second degree polynomial, where x is the amount 
invested in thousands of dollars and R(x) is the amount of revenue earned by the firm (also in 
thousands of dollars). 



R(x) 



-4.1a;^ + 166.8a; + 6196 



Use the graph and then the polynomial to estimate the firm's revenue when the firm invested 
$10,000 in advertising. 

R (thousands of dollars) 

8,500 



8,000 
7,500 
7,000 
6,500 
6,000 



(i 

.A it 

> 



x (thousands of dollars) 



5 10 15 20 25 30 

2. The table below lists the estimated number of aids cases in the United States for the years 1999- 
2003. 



Year 



AIDS Cases 



1999 



2000 



2002 



2003 



41,356 41,267 41,289 43,171 



The data is plotted then fitted with the following second degree polynomial, where t is the number 
of years that have passed since 1998 and N(t) is the number of aids case reported t years after 
1998. 

N(t) = 345. 14i 2 - 1705.7* + 42904 



Use the graph and then the polynomial to estimate the number of AIDS cases in the year 2001. 



5.3. APPLICATIONS OF POLYNOMIALS 



331 



N (Number of AID cases) 
45,000 



44,000 
43,000 
42,000 
41,000 
40,000 







1 



t (Years since 1998) 



3. The following table records the concentration (in milligrams per liter) of medication in a patient's 
blood after indicated times have passed. 



Time (Hours) 





0.5 


1 


1.5 


2.5 


Concentration (mg/L) 





78.1 


99.8 


84.4 


15.6 



The data is plotted then fitted with the following second degree polynomial, where t is the number 
of hours that have passed since taking the medication and C(t) is the concentration (in milligrams 
per liter) of the medication in the patient's blood after t hours have passed. 



C(t) 



-56.214r + 139.3K + 9.35 



Use the graph and then the polynomial to estimate the the concentration of medication in the 
patient's blood 2 hours after taking the medication. 
C (mg/L) 



, 
























i 


1 \ 






oU " 


( 


( / 










DU " 














4U " 














zu ■ 










\« 


t 


n i 


1 1 










\ 



-> t (Hours) 



332 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



4. The following table records the population (in millions of people) of the United States for the 
given year. 



Year 


1900 


1920 


1940 


1960 


1980 


2000 


2010 


Population (millions) 


76.2 


106.0 


132.2 


179.3 


226.5 


281.4 


307.7 



The data is plotted then fitted with the following second degree polynomial, where t is the number 
of years that have passed since 1990 and P(t) is the population (in millions) t years after 1990. 

P(t) = 0.008597* 2 + 1, 1738i + 76.41 

Use the graph and then the polynomial to estimate the the population of the United States in 
the year 1970. 

P (millions of people) 




20 40 60 80 100 120 



t (years since 1990) 



If a projectile is launched with an ini- 
tial velocity of 457 meters per second 
(457 m/s) from a rooftop 75 meters (75 m) 
above ground level, at what time will the 
projectile first reach a height of 6592 me- 
ters (6592 m)? Round your answer to 
the nearest second. Note: The accelera- 
tion due to gravity near the earth's sur- 
face is 9.8 meters per second per second 
(9.8 m/s 2 ). 

If a projectile is launched with an ini- 
tial velocity of 236 meters per second 
(236 m/s) from a rooftop 15 meters (15 m) 



above ground level, at what time will the 
projectile first reach a height of 1838 me- 
ters (1838 m)? Round your answer to 
the nearest second. Note: The accelera- 
tion due to gravity near the earth's sur- 
face is 9.8 meters per second per second 
(9.8 m/s 2 ). 

If a projectile is launched with an ini- 
tial velocity of 229 meters per second 
(229 m/s) from a rooftop 58 meters (58 m) 
above ground level, at what time will the 
projectile first reach a height of 1374 me- 
ters (1374 m)? Round your answer to 



5.3. APPLICATIONS OF POLYNOMIALS 



333 



the nearest second. Note: The accelera- 
tion due to gravity near the earth's sur- 
face is 9.8 meters per second per second 
(13.8 m/s 2 ). 

If a projectile is launched with an ini- 
tial velocity of 234 meters per second 
(234 m/s) from a rooftop 16 meters (16 m) 



above ground level, at what time will the 
projectile first reach a height of 1882 me- 
ters (1882m)? Round your answer to 
the nearest second. Note: The accelera- 
tion due to gravity near the earth's sur- 
face is 9.8 meters per second per second 
(9.8 m/s 2 ). 



In Exercises 9-12, first use an algebraic technique to find the zero of the given function, then use 
the 2:zero utility on your graphing calculator to locate the zero of the function. Use the Calculator 
Submission Guidelines when reporting the zero found using your graphing calculator. 



9. f{x) = 3.25a; - 4.875 
10. f(x) = 3.125 -2.5a; 



11. f(x) = 3.9- 1.5a; 

12. f(x) = 0.75a; + 2.4 



13. If a projectile is launched with an ini- 
tial velocity of 203 meters per second 
(203 m/s) from a rooftop 52 meters (52 m) 
above ground level, at what time will the 
projectile return to ground level? Round 
your answer to the nearest tenth of a sec- 
ond. Note: The acceleration due to grav- 
ity near the earth's surface is 9.8 meters 
per second per second (9.8 m/s 2 ). 

14. If a projectile is launched with an ini- 
tial velocity of 484 meters per second 
(484 m/s) from a rooftop 17 meters (17 m) 
above ground level, at what time will the 
projectile return to ground level? Round 
your answer to the nearest tenth of a sec- 
ond. Note: The acceleration due to grav- 
ity near the earth's surface is 9.8 meters 
per second per second (9.8 m/s 2 ). 



15. If a projectile is launched with an ini- 
tial velocity of 276 meters per second 
(276 m/s) from a rooftop 52 meters (52 m) 
above ground level, at what time will the 
projectile return to ground level? Round 
your answer to the nearest tenth of a sec- 
ond. Note: The acceleration due to grav- 
ity near the earth's surface is 9.8 meters 
per second per second (9.8 m/s 2 ). 

16. If a projectile is launched with an ini- 
tial velocity of 204 meters per second 
(204 m/s) from a rooftop 92 meters (92 m) 
above ground level, at what time will the 
projectile return to ground level? Round 
your answer to the nearest tenth of a sec- 
ond. Note: The acceleration due to grav- 
ity near the earth's surface is 9.8 meters 
per second per second (9.8 m/s 2 ). 



;*.;-». j». 



Answers ■** -** •** 



1. Aprpoximately $7,452 
3. Approximately 63 mg/L 
5. 17.6 seconds 
7. 6.7 seconds 



9. Zero: 1.5 
11. Zero: 2.6 
13. 41.7 seconds 
15. 56.5 seconds 



334 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Simplify: 

(3s 2 -2si + 4i 2 ) + 
(s 2 + 7 at - 5t 2 ) 



Answer: 4s 2 + 5st — t 2 



5.4 Adding and Subtracting Polynomials 

In this section we concentrate on adding and subtracting polynomial expres- 
sions, based on earlier work combining like terms in Ascending and Descending Powers 
on page 309. Let's begin with an addition example. 



EXAMPLE 1. Simplify: 

(a 2 + 3ab - b 2 ) + (4a 2 + llab - 9b 2 ) 

Solution: Use the commutative and associative properties to change the order 
and regroup. Then combine like terms. 

(a 2 + 3ab - b 2 ) + (4a 2 + llab - 96 2 ) 

= (a 2 + 4a 2 ) + (3a6 + llab) + (-b 2 - 9b 2 ) 
= 5a 2 + Uab- 10b 2 



a 



Let's combine some polynomial functions. 



You Try It! 



Given f(x) = 2x 2 + 9x - 5 



and g(x) 



4a; + 3, 



simplify f(x) + g(x). 



Answer: x 2 + 5x — 2 



EXAMPLE 2. Given f(x) =3x 2 -4x-8 and g(x) = x 2 - lis + 15, simplify 

f(x)+g(x). 

Solution: First, replace fix) and g{x) with their definitions. Be sure to 
surround each polynomial with parentheses, because we are asked to add all 
of f{x) to all of g(x). 

f{x) + g(x) = (3a; 2 - 4sc - 8) + (a; 2 - 11a; + 15) 

Now use the commutative and associative properties to change the order and 
regroup. Combine like terms. 

= (3a; 2 + a; 2 ) + (-4a; - 11a;) + (-8 + 15) 
= 4a; 2 - 15a: + 7 

Hence, fix) + gix) = 4a; 2 — 15a; + 7. 
□ 

If you are comfortable skipping a step or two, it is not necessary to write 
down all of the steps shown in Examples 1 and 2. Let's try combining like 
terms mentally in the next example. 



5.4. ADDING AND SUBTRACTING POLYNOMIALS 



335 



EXAMPLE 3. Simplify: 

(a; 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + by 3 ) 



You Try It! 



Simplify: 

(-ba 2 b + 4ab - 3ab 2 )- 
{2a 2 b + Tab - ab 2 ) 



Solution: If we use the associative and commutative property to reorder and 
regroup, then combine like terms, we get the following result. 

(x 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + 5y 3 ) 

= {x 3 + 2x 3 ) + {-2x 2 y - 4x 2 y) + (3a;?/ 2 - 8xy 2 ) + (y 3 + by 3 ) 
= 2.x 3 - 6x 2 y - 5xy 2 + Qy 3 

However, if we can combine like terms mentally, eliminating the middle step, 
it is much more efficient to write: 

(a; 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + by 3 ) 
= 3x — 6x y — bxy + 6y 



Answer: 
-3a 2 b+lla6-4a6 2 d 



a 



Negating a Polynomial 

Before attempting subtraction of polynomials, let's first address how to negate 
or "take the opposite" of a polynomial. First recall that negating is equivalent 
to multiplying by — 1 . 




We can use this property to simplify — (a + b). First, negating is identical to 
multiplying by —1. Then we can distribute the — 1. 

— (a + b) = (— l)(a + b) Negating is equivalent to multiplying 

by-1. 

= (-l)a + (-1)6 Distribute the -1. 

= —a + (—b) Simplify: (— l)a = —a and (—1)6 = — b. 

= —a — b Subtraction means add the opposite. 



336 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Thus, — (a + b) = —a — b. However, it is probably simpler to note that the 
minus sign in front of the parentheses simply changed the sign of each term 
inside the parentheses. 



Negating a sum. When negating a sum of terms, the effect of the minus sign 
is to change each term in the parentheses to the opposite sign. 



-(a + < 



-a — b 



You Try It! 



Simplify: -(2a; 2 - 3a; + 9) 



Let's look at this principle in the next example. 



EXAMPLE 4. Simplify: -(-3a; 2 + 4a; - 8) 

Solution: First, negating is equivalent to multiplying by —1. Then distribute 
the -1. 



(-3a; 2 + 4a; -8) 
= (-l)(-3a; 2 + 4a;- 



Negating is equivalent to 
multiplying by —1. 



(-l)(-3a; 2 ) + (-l)(4a;) - (-1)(8) Distribute the -1. 



3a; 2 + (-4a;) - (- 



3a; 2 - 4x + : 



Simplify: (-l)(-3a; 2 ) = 3a; 2 
(-l)(4a;) = -4a;, and 
and (-1)(8) = -8. 

Subtraction means add 
the opposite. 



Answer: — 2x 2 + 3a; — 9 



Alternate solution: As we saw above, a negative sign in front of a parentheses 
simply changes the sign of each term inside the parentheses. So it is much more 
efficient to write 



-(-3a; 2 +4a;- 



3a; 2 - 4a; + I 



simply changing the sign of each term inside the parentheses. 



□ 



Subtracting Polynomials 

Now that we know how to negate a polynomial (change the sign of each term 
of the polynomial), we're ready to subtract polynomials. 



5.4. ADDING AND SUBTRACTING POLYNOMIALS 



337 



You Try It! 



EXAMPLE 5. Simplify: 

(y 3 - 3y 2 z + Ayz 2 + z 3 ) - (2y 3 - 8y 2 z + 2yz 2 - 8z 3 ) 

Solution: First, distribute the minus sign, changing the sign of each term of 
the second polynomial. 

[y 3 - 3y 2 z + Ayz 2 + z 3 ) - {2y 3 - 8y 2 z + 2yz 2 - 8z 3 ) 
= y 3 - 3y 2 z + Ayz 2 + z 3 - 2y 3 + 8y 2 z - 2yz 2 + 8z 3 

Regroup, combining like terms. You may perform this next step mentally if 
you wish. 

= (y 3 - 2y 3 ) + {-3y 2 z + 8y 2 z) + (Ayz 2 - 2yz 2 ) + (z 3 + 8z 3 ) 
= -y 3 + 5y 2 z + 2yz 2 + 9z 3 



Simplify: 

(Aa 2 b + 2ab-7ab 2 )- 
(2a 2 b -ab- 5ab 2 ) 



Answer: 2a 2 b + 3ab - 2ab 2 



D 



Let's subtract two polynomial functions. 



You Try It! 



6ar 



7x- 11, 



EXAMPLE 6. Given p(x) = -5x 3 + 6x - 9 and q(x) 
simplify p(x) — q(x). 

Solution: First, replace p(x) and q(x) with their definitions. Because we are 
asked to subtract all of q(x) from all of p(x), it is critical to surround each 
polynomial with parentheses. 

p{x) - q(x) = {-5x 3 + 6x - 9) - (6x 2 - 7x - 11) 

Distribute the minus sign, changing the sign of each term in the second poly- 
nomial, then regroup and combine like terms. 

= -5x 3 + 6x - 9 - 6a; 2 + 7x + 11 

= -5x 3 - 6x 2 + (6x + 7x) + (-9 + 11) 

= -5x 3 -6a; 2 + 13a: + 2 

However, after distributing the minus sign, if we can combine like terms men- 
tally, eliminating the middle step, it is much more efficient to write: 

p(x) - q(x) = (-5a; 3 + 6x - 9) - (6a; 2 - 7x - 11) 
= -5a; 3 + 6a; - 9 - 6a; 2 + 7a; + 11 
= -5a; 3 - 6a; 2 + 13a; + 2 



Given f(x) = 3x 2 + 9x - 4 
and g(x) = — 5x 2 + Ax — 6, 
simplify f(x) - g(x). 



Answer: 8a; 2 + 5a; + 2 



□ 



338 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Some Applications 

Recall that the area of a rectangle having length L and width W is found using 
the formula A = LW . The area of a square having side s is found using the 
formula A = s 2 (see Figure 5.31). 



W 



A = LW 




L 



Figure 5.31: Area formulae for the rectangle and square. 



You Try It! 



Find the area of the square 
shown below by summing 
the area of its parts. 
x 4 



EXAMPLE 7. Find the area of the square in Figure 5.32 by summing the 
area of its parts. 

Solution: Let's separate each of the four pieces and label each with its area 
(see Figure 5.33). 





X 


3 


3 






X 







Figure 5.32: Find the sum of the 
parts. 



A 2 = 3x 






Figure 5.33: Finding the area of 
each of the four parts. 



The two shaded squares in Figure 5.33 have areas A\ = x 2 and A% = 9, respec- 
tively. The two unshaded rectangles in Figure 5.33 have areas A 2 = 3x and 
A4 = 3x. Summing these four areas gives us the area of the entire figure. 



Answer: 



8a; +16 



3x + 9 + 3x 
6x + 9 



□ 



5.4. ADDING AND SUBTRACTING POLYNOMIALS 



339 



You Try It! 



EXAMPLE 8. Ginger runs a business selling wicker baskets. Her business 
costs for producing and selling x wicker baskets are given by the polynomial 
function C(x) = 100 + 3a; — 0.02a; 2 . The revenue she earns from selling x wicker 
baskets is given by the polynomial function R(x) = 2.75a;. Find a formula for 
P{x), the profit made from selling x wicker baskets. Use your formula to 
determine Ginger's profit if she sells 123 wicker baskets. 

Solution: The profit made from selling x wicker baskets is found by subtract- 
ing the costs incurred from the revenue received. In symbols: 

P{x) = R(x) - C(x) 

Next, replace R(x) and C(x) with their definitions. Because we are supposed 
to subtract all of the cost from the revenue, be sure to surround the cost 
polynomial with parentheses. 

P(x) = 2.75a; - (100 + 3a; - 0.02a; 2 ) 

Distribute the minus sign and combine like terms. 

= 2.75a;- 100 - 3x + 0.02a; 2 
= 0.02a; 2 - 0.25a; - 100 

Thus, the profit function is P{x) = 0.02a; 2 - 0.25a; - 100. 

Next, to determine the profit if 123 wicker baskets are sold, substitute 123 
for x in the profit function P{x). 

P(x) = 0.02a; 2 - 0.25a; - 100 
P(123) = 0.02(123) 2 - 0.25(123) - 100 

You can now use your graphing calculator to determine the profit (see Figure 5.34). 
Hence, the profit made from selling 123 wicker baskets is $171.83. 



The costs for producing and 
selling x widgets are given 
by the polynomial function 
C(x) =50 + 5a; -0.5a; 2 , and 
the revenue for selling x 
widgets is given by the 
polynomial function 
R{x) = 3.5a;. Determine the 
profit if 75 widgets are sold. 



0.02*123^2-0.25* 
123-100 

171. S3 



Figure 5.34: Determining the profit from selling 123 wicker baskets. 



Answer: $2,650 



□ 



340 CHAPTER 5. POLYNOMIAL FUNCTIONS 

**> t* t* Exercises ■** ■*$ •*$ 

In Exercises 1-8, simplify the given expression. Arrange your answer in some sort of reasonable order. 

1. (-8r 2 t + 7rt 2 + 3t 3 ) + (9r 3 + 2rt 2 + At 3 ) 5. (-2r 2 + 7rs + 4s 2 ) + (-9r 2 + 7rs - 2s 2 ) 

2. (-a 3 - 8ac 2 - 7c 3 ) + (-7a 3 - 8a 2 c + 8ac 2 ) 6. (-2r 2 + 3rt - 4i 2 ) + (7r 2 + Art - It 2 ) 

3. (7x 2 - 6x - 9) + (8a; 2 + 10sc + 9) 7. (-8y 3 - 3y 2 z - 6z 3 ) + (-3y 3 + 7y 2 z - 9yz 2 ) 

4. (-7x 2 + 5x - 6) + (-10s 2 - 1) 8. {7y 2 z + 8yz 2 + 2z 3 ) + (8y 3 - 8y 2 z + 9yz 2 ) 

In Exercises 9-14, simplify the given expression by distributing the minus sign. 

9. -(5x 2 - 4) 12. -(7w 3 - 8u 2 v + 6uv 2 + 5v 3 ) 

10. -(-8a; 2 -5) 13. ~{-5x 2 + 9xy + 6y 2 ) 

11. -(9r 3 - Ar 2 t - 3rt 2 + At 3 ) 14. -(-4u 2 - Guv + 5v 2 ) 

In Exercises 15-22, simplify the given expression. Arrange your answer in some sort of reasonable order. 

15. (-u 3 - Au 2 w + 7w 3 ) - (u 2 w + uw 2 + 3w 3 ) 19. (-7r 2 - 9rs - 2s 2 ) - (-8r 2 - 7rs + 9s 2 ) 

16. (-b 2 c + 86c 2 + 8c 3 ) - (6b 3 + b 2 c - Abe 2 ) 20. (-4a 2 + bab - 2b 2 ) - (-8a 2 + 7a6 + 26 2 ) 

17. (2y 3 - 2y 2 2 + 3z 3 ) - (-8y 3 + hyz 2 - 3z 3 ) 21. (10a; 2 + 2x - 6) - (-8a; 2 + 14a; + 17) 

18. (4a 3 + 6ac 2 + 5c 3 ) - (2a 3 + 8a 2 c - 7ac 2 ) 22. (-5a; 2 + 19a; - 5) - (-15a; 2 + 19a; + 8) 



In Exercises 23-28, for the given polynomial functions f(x) and g(x), simplify f(x) + g(x). Arrange 
your answer in descending powers of x. 

23. f(x) = -2x 2 +9x + 7 26. 



24. f(x) = -8x 3 +6x-9 27. 



/(*) = 


■■ -2a; 2 + 9a; + 7 


g(x) = 


: 8a; 3 - 7a; 2 + 5 


m = 


: -8x 3 +6a;-9 


g(x) = 


: a; 3 - a; 2 + 3a; 


/(*) = 


: 5a; 3 - 5a; 2 + 8a; 


g(x) = 


: 7a; 2 - 2a; - 9 



25. f(x) = bx° -bx* + 8x 28. 



m = 


-a; 2 - 


f 8a; +1 


g(x) = 


-7a; 3 


+ 8a;-9 


m = 


-3a; 2 


-8a; -9 


g(x) = 


5a; 2 - 


- 4a; + 4 


/(*) = 


-3a; 2 


+ a;-8 



g(x) = 7a; 2 - 9 



5.4. ADDING AND SUBTRACTING POLYNOMIALS 



341 



In Exercises 29-34, for the given polynomial functions f(x) and g(x), simplify f(x) — g(x). Arrange 
your answer in descending powers of x. 



29. 



30. 



31. 



9{x) 



-6x J 



7a; + 7 



-3a; 3 - 3x z 



8x 



f( x ) = 5a; 3 - 5x + 4 
g (x) = -8x 3 - 2x 2 - 

f(x) = 12a; 2 -5a; + 4 
g(x) = 8x 2 - 16a;- 7 



3x 



32. 



33. 



34. 



m = 


-7x 2 + Ylx + 17 


g{x) = 


-10a; 2 - 17 


f{x) = 


-3a; 3 - 4a; + 2 


g(x) = 


-4a; 3 - 7a; 2 + 6 


M = 


-9a; 2 + 9a; + 3 




7^,3 | njl | e 



In Exercises 35-36, find the area of the given square by summing the areas of its four parts. 
35. 36. 





X 


5 


5 






X 









X 


7 


7 






X 







37. Rachel runs a small business selling wicker 
baskets. Her business costs for produc- 
ing and selling x wicker baskets are given 
by the polynomial function C(x) = 232 + 
7a;— 0.0085a; 2 . The revenue she earns from 
selling x wicker baskets is given by the 
polynomial function R(x) = 33.45a;. Find 
a formula for P{x), the profit made from 
selling x wicker baskets. Use your formula 
to determine Rachel's profit if she sells 233 
wicker baskets. Round your answer to the 
nearest cent. 



38. Eloise runs a small business selling baby 
cribs. Her business costs for producing 
and selling x baby cribs are given by the 
polynomial function C(x) = 122 + 8a; — 
0.0055a; 2 . The revenue she earns from sell- 
ing x baby cribs is given by the polyno- 
mial function R(x) = 33.45a;. Find a for- 
mula for P(x), the profit made from sell- 
ing x baby cribs. Use your formula to de- 
termine Eloise's profit if she sells 182 baby 
cribs. Round your answer to the nearest 
cent. 



342 CHAPTER 5. POLYNOMIAL FUNCTIONS 

£»- j*- &*■ Answers •*$ *** •** 

1. 9r 3 - 8r 2 t + 9rt 2 + 7£ 3 21. 18x 2 - 12a; - 23 

3. 15a; 2 + 4a; 23. 8a; 3 - 9a; 2 + 9x + 12 

5. -llr 2 + 14rs + 2s 2 r ^3 , o„2 



7. -lly 3 + 4y 2 z - 9yz 2 - 6z 3 

9. -5a; 2 + 4 
11. -9r 3 + 4r 2 t + 3r£ 2 - At 3 
13. 5a; 2 — 9a;y — 6y 2 
15. — u 3 — 5u 2 w — uw 2 + Aw 3 
17. 10y 3 - 2y 2 z - 5yz 2 + 6z 3 
19. r 2 - 2rs- lis 2 



25. 5x 3 + 2aT + 6a; - 9 
27. 2a; 2 - 12a; - 5 
29. -3a; 3 + 3a; 2 + x + 7 
31. 4x 2 + lla;+ll 
33. a; 3 + 7a; 2 - 4a; - 4 
35. a; 2 + 10a; + 25 
37. $6,392.31 



5.5. LAWS OF EXPONENTS 



343 



5.5 Laws of Exponents 

In Chapter 1, section 1, we first introduced the definition of an exponent. For 
convenience, we repeat that definition. 

In the exponential expression a™, the number a is called the base, while the 
number n is called the exponent. 

Exponents. Let a be any real number and let n be any whole number. If 
ra^O, then: 

a n = a ■ a ■ a a 



That is, to calculate a", write a as a factor n times. In the case where a/0, 
but n = 0, then we define: 



o° = l 



For example, raising a number to the fifth power requires that we repeat the 
number as a factor five times (see Figure 5.35). 

(_2) 5 = (-2)(-2)(-2)(-2)(-2) 
= -32 

Raising a number to the fourth power requires that we repeat that number as 
a factor four times (see Figure 5.35). 



1 H H H 



i 

16 



As a final example, note that 10° = 1, but 0° is undefined (see Figure 5.36). 



<-l/-2) A 4 
RnsN-Frac 



-32 

.0625 

1/16 



Figure 5.35: Evaluating (—2) and 
(-1/2) 4 . Recall: Use the MATH 
key to locate the ►Frac command. 



10^0 
0^0 



1 

Error 



Figure 5.36: Evaluating 10° and 0° 
on the graphing calculator. 



344 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



For those who may be 
wondering why a = 1, 
provided a ^ 0, here is a nice 
argument. First, note that 



a-a° = a 1 -a° 



On the right, repeat the base 
and add the exponents. 



a ■ a = a 



Or equivalently: 



Multiplying With Like Bases 

In the expression a n , the number a is called the base and the number n is 
called the exponent. Frequently, we'll be required to multiply two exponential 
expressions with like bases, such as x 3 • x 4 . Recall that the exponent tells us 
how many times to write each base as a factor, so we can write: 

x 3 ■ x = (x ■ x ■ x) ■ (x ■ x ■ x ■ x) 

tJU *Xj tJU Ju Ju t*> iA; 

= x 7 

Note that we are simply counting the number of times that x occurs as a factor. 
First, we have three it's, then four cc's, for a total of seven x J s. However a little 
thought tells us that it is much quicker to simply add the exponents to reveal 
the total number of times x occurs as a factor. 



a ■ a = a 



3 4 
X ■ X 



„3+4 



Now, divide both sides by a, 
which is permissable if a 7^ 0. 



a ■ a a 
a a 



Simplify both sides: 



The preceding discussion is an example of the following general law of expo- 
nents. 



Multiplying 


With Like Bases. 


To multiply two exponential 


expressions 


with like bases 


, repeat the base and add the exponents. 






m 

a 


n m-\-n 

a = a 





You Try It! 



Simplify: 3 4 • 3 2 



EXAMPLE 1. Simplify each of the following expressions: 
(a)y 4 V (b)2 3 -2 5 {c)ix + y) 2 {x + y y 

Solution: In each example we have like bases. Thus, the approach will be the 
same for each example: repeat the base and add the exponents. 

(a) y A ■ y s = y i+s (b) 2 3 • 2 5 = 2 3 + 5 (c) (x + yf{x + y) 7 = (x + yf +7 



Answer: 3 6 



y 



12 



{x + yf 



With a little practice, each of the examples can be simplified mentally. Repeat 



the base and add the exponents in your head: y ■ y 

(x + y) 2 (x + y) 7 = {x + yf. 



V 



2 A ■ 2 b 



2 8 , and 



□ 



5.5. LAWS OF EXPONENTS 



345 



You Try It! 



EXAMPLE 2. Simplify: (a 6 & 4 )(a 3 6 2 ) 

Solution: We'll use the commutative and associative properties to change the 
order of operation, then repeat the common bases and add the exponents. 



Simplify: (x y )(x y ) 



3(,2n 



(a b b 4 ){a 3 b 



a 6 b 4 a 3 b 2 
a 6 a 3 b 4 b 2 



9 7,6 



a y 6 



The associative property allows us 
to regroup in the order we prefer. 

The commutative property allows us 
to change the order of multiplication. 

Repeat the common bases and add the 
exponents. 



With practice, we realize that if all of the operators are multiplication, then we 
can multiply in the order we prefer, repeating the common bases and adding 



the exponents mentally: (a b )(a 



,3/,2n. 



J 6 6 . 



Answer: x 6 y 9 



□ 



You Try It! 



EXAMPLE 3. Simplify: x n+3 ■ x 3 ~ 2n 

Solution: Again, we repeat the base and add the exponents. 

; (n+3)+(3-2n) R epea t the base, add the exponents. 



Simplify: x 



5—n , 4n+2 



x n+3 ■ x 3 ~ 2n 



„6— n 



Simplify. Combine like terms. 



Answer: x 3n+r 



Dividing With Like Bases 

Like multiplication, we will also be frequently asked to divide exponential ex- 
pressions with like bases, such as x 7 /x A . Again, the key is to remember that 
the exponent tells us how many times to write the base as a factor, so we can 
write: 



□ 



X 


t*> iXj *Xj *AJ iJU *JU 


X 


x ■ x ■ x ■ x- 

r p ■ ht ' It ' T* ■ If* ■ If* 



x'-x'-x'-x'- 



Note how we cancel four x's in the numerator for four x's in the denominator. 
However, in a sense we are "subtracting four x's" from the numerator, so a 



346 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Here is another nice 
argument why a° = 1, 
provided a/0. Start with: 



faster way to proceed is to repeat the base and subtract the exponents, as 
follows: 



„7-4 



1 



Repeat the base and 
subtract the exponents. 



a 1 - 1 = 1 



Simplify. 



The preceding discussion is an example of the second general law of expo- 
nents. 



Dividing With Like Bases. To divide two exponential expressions with like 
bases, repeat the base and subtract the exponents. Given a/0, 



You Try It! 



4 5 
Simplify: — 



Note that the subtraction of the exponents follows the rule "top minus bottom." 



EXAMPLE 4. Simplify each of the following expressions: 



™12 

X 6 



(b)lf 



(c) 



(2x+l) 8 
(2a; + 1) 3 



Solution: In each example we have like bases. Thus, the approach will be the 
same for each example: repeat the base and subtract the exponents. 



( a )^T 



„12-3 






5 7 - 7 

5° 

1 



(c) 



(2x + 1) 8 
(2x + 1) 3 



(2s + 1) 
(2s + 1) 



8-3 



Answer: 4 2 



With a little practice, each of the examples can be simplified mentally. Repeat 
the base and subtract the exponents in your head: x 12 /x 3 = x 9 , 5 7 /5 4 = 5 3 , 
and (2x + l) 8 /(2ir + l) 3 = (2x + l) 5 . 

D 



You Try It! 



Simplify: 



15a 6 6 9 
3a6 5 



5, ,7 



EXAMPLE 5. Simplify: t X „\ 

Solution: We first express the fraction as a product of three fractions, the 
latter two with a common base. In the first line of the following solution, 



5.5. LAWS OF EXPONENTS 



347 



note that if you multiply numerators and denominators of the three separate 
fractions, the product equals the original fraction on the left. 



5„,7 



12 x 5 y 7 



j.3 y2 



I2x*y 

4 X 3y2 4 

= 3x 5 - V- 2 
= 3xV 



Break into a product of three fractions. 

Simplify: 12/4=3. Then repeat the common 
bases and subtract the exponents. 

Simplify. 



Answer: 5a 5 6 4 



□ 



You Try It! 



„5n— 4 



EXAMPLE 6. Simplify: 



»3— 2n 

Solution: Again, we repeat the base and subtract the exponents. 

„5n— 4 



Simplify: 



~3n— 6 



„n+2 



;)'■ 



3-2n 



„(5n-4)-(3-2n) 



„5n-4-3+2ra 



„7n-7 



Repeat the base, subtract exponents. 

Distribute the minus sign. 
Simplify. Combine like terms. 



Answer: 



n 2n-8 



Raising a Power to a Power 

Suppose we have an exponential expression raised to a second power, such as 
(a; 2 ) 3 . The second exponent tells us to write x 2 as a factor three times: 



l 2\3 2 2 

(X ) = X ■ X ■ X 

„6 



Write x as a factor three times. 
Repeat the base, add the exponents. 



Note how we added 2 + 2 + 2 to get 6. However, a much faster way to add 
"three twos" is to multiply: 3-2 = 6. Thus, when raising a "power to a second 
power," repeat the base and multiply the exponents, as follows: 

(x 2 ) 3 = x 2 - 3 

™6 



□ 



The preceding discussion gives rise to the following third law of exponents. 



348 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Answer: a 



6— 3n 



Raising a Power to 


a Power. 


When raising 


a power 


to 


a power, 


repeat the 


base and multiply the exponents 


. In symbols: 














(a m ) n = a mn 










Note that juxtaposing 


two variables, as in inn, 


means 


'm 


times n.' 


5 



You Try It! 



Simplify: (2 



3\4 



EXAMPLE 7. Simplify each of the following expressions: 

(a) (z 3 ) 5 (b) (7 3 ) (c) [(a; - y) 3 ] 6 

Solution: In each example we are raising a power to a power. Hence, in each 
case, we repeat the base and multiply the exponents. 



3\5 _ „3-5 



(a) (zy = z 



(b) (7 



3\0 _ 7 3-0 



(c) [(x - y) 



3 ifi 



v lo 



(x-yf 6 

(x-y) 18 



Answer: 2 12 



With a little practice, each of the examples can be simplified mentally. Repeat 



the base and multiply the exponents in your head: (z 
and [(x — y) 3 ] 6 = (x — y) ls - 



3\5 _ ,15 (^4 



(7 3 



7 



12 



□ 



You Try It! 



Simplify: (a 



2-n\3 



„2n-3\4 



EXAMPLE 8. Simplify: 

Solution: Again, we repeat the base and multiply the exponents. 



(,: 



2n-3\4 _ 4(2n-3) 



Repeat the base, multiply exponents. 
Distribute the 4. 



□ 



5.5. LAWS OF EXPONENTS 



349 



Raising a Product to a Power 

We frequently have need to raise a product to a power, such as {xy) 3 . Again, 
remember the exponent is telling us to write xy as a factor three times, so: 



to/) 3 



{xy){xy){xy) 
xyxyxy 

xxxyyy 



x 3 y 3 



Write xy as a factor three times. 

The associative property allows us to 
group as we please. 

The commutative property allows us to 
change the order as we please. 

Invoke the exponent definition: 
xxx = x and yyy = y . 



However, it is much simpler to note that when you raise a product to a power, 
you raise each factor to that power. In symbols: 



{xy? 



x 3 y 3 



The preceding discussion leads us to a fourth law of exponents. 



Raising a P 


roduct to 


a Power. To 


raise a 


product 


to 


a 


power, 


raise 


each 


factor 


to that 


power. 


In 


symbols: 

(ab) n = 


--a n b n 















EXAMPLE 9. Simplify each of the following expressions: 

(a) {yzf (b) {-2xf (c) (-3y) 2 

Solution: In each example we are raising a product to a power. Hence, in 
each case, we raise each factor to that power. 



You Try It! 



Simplify: {-2b) 4 



5,5 



(a) {yzf = y b z 



(b) (-2a;) 3 = (-2) 3 a; 3 
= -8x 3 



(c) {-3yf = (-3)V 
= 9y 2 



With a little practice, each of the examples can be simplified mentally. Raise 
each factor to the indicated power in your head: {yzf = y 5 z 5 , (—2a;) 3 = —8a; 3 , 
and (-3y) 2 = 9y 2 



Answer: 166 



When raising a product of three factors to a power, it is easy to show that we 



should raise each factor to the indicated power. For example, (abcf 



3 6 3 c 



In general, this is true regardless of the number of factors. When raising a 
product to a power, raise each of the factors to the indicated power. 



□ 



350 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Simplify: (—3xy 



4\5 



EXAMPLE 10. Simplify: (-2a 3 6 2 ) 3 

Solution: Raise each factor to the third power, then simplify. 

(-2a 3 6 2 ) 3 = (-2) 3 (a 3 ) 3 (6 2 ) 3 Raise each factor to the third power. 

= — 8a 9 & 6 Simplify: (— 2) 3 = 8. In the remaining factors, 



raising a power to a power requires that 
we multiply the exponents. 



Answer: — 243xy 



5, ,20 



□ 



You Try It! 



Simplify: (—a 



3/,2\3/ 



-2a 2 b 



2ui\2 



Answer: — 4o b 



131,14 



EXAMPLE 11. Simplify: (-2x 2 y) 2 {-3x 3 y) 

Solution: In the first grouped product, raise each factor to the second power. 



(-2x 2 y) 2 (-3x 3 y) = ((-2) 2 (x 2 ) 2 y 2 )(-3x 3 y) 



(Ax A y 2 )(-3x 3 y) 



Raise each factor in the first 
grouped product to the second 
power. 



Simplify: 
(x 2 ) 2 =x 



4 and 



The associative and commutative property allows us to multiply all six factors 
in the order that we please. Hence, we'll multiply 4 and —3, then x 4 and x 3 , 
and y 2 and y, in that order. In this case, we repeat the base and add the 
exponents. 



-12xV 



Simplify: (4)(-3) = 
x 7 and y 2 y 



x 4 x 3 



-12. Also, 

.,3 



□ 



Raising a Quotient to a Power 

Raising a quotient to a power is similar to raising a product to a power. For 
example, raising (x/y) 3 requires that we write x/y as a factor three times. 

\ 3 
x 

I) 



X 


X X 


y 


y y 


X 


X ■ X 



vvv 

^■3 



5.5. LAWS OF EXPONENTS 



351 



However, it is much simpler to realize that when you raise a quotient to a 
power, you raise both numerator and denominator to that power. In symbols: 

x\ x 3 

,yj V 3 

This leads to the fifth and final law of exponents. 



Raising a Quotient to a Power. To raise a quotient to a power, raise both 
numerator and denominator to that power. Given 6^0, 



(a\ n a n 
\b ) ~ 6" 



EXAMPLE 12. Simplify each of the following expressions: 
' : >>(^) 2 (V(~Y (c) 



Solution: In each example we are raising a quotient to a power. Hence, in 
each case, we raise both numerator and denominator to that power. 



You Try It! 



Simplify: 



""!§ 



22 

3 2 

4 

9 



<«(f)H 

X 3 

~ 27 



(c) 



24 

16 



Note that in example (c), raising a negative base to an even power produces a 
positive result. With a little practice, each of the examples can be simplified 
mentally. Raise numerator and denominator to the indicated power in your 
head: (2/3) 2 = 4/9, (x/3) 3 = £ 3 /27, and (-2/j/) 4 = 16/t/ 4 



Answer: 



125 
"64" 



□ 



EXAMPLE 13. 

Simplify: I — 5- 
V V 

Solution: Raise both numerator and denominator to the second power, then 

simplify. 

2 '~„5\2 



2a; 5 

y 3 



(2x 5 f 

(y 3 ) 2 



Raise numerator and denominator to the 
second power. 



You Try It! 



Simplify: ( — 



^ 

Answer: ^ 



352 CHAPTER 5. POLYNOMIAL FUNCTIONS 

In the numerator, we need to raise each factor of the product to the second 
power. Then we need to remind ourselves that when we raise a power to a 
power, we multiply the exponents. 



2/„.5\2 



2 2 (x 5 ) 

(y 3 ) 2 



Raise each factor in the numerator 
to the second power 



l2 a l„b\2 _ 10 



Simplify: 2 Z =4, (x"Y = x l 
V and (y 3 ) 2 = y e . 



D 



5.5. LAWS OF EXPONENTS 353 

**. **. **. Exercises -*s -« >•$ 

In Exercises 1-8, simplify each of the given exponential expressions. 

1. (-4) 3 

2. (-9) 2 

o 



3. 



4. 



5 
o 



5. 


B) 2 


6. 


(-!)" 


7. 


(-19)0 


8. 


(-17)0 



In Exercises 9-18, simplify each of the given exponential expressions. 

9. (7v - 6w) 18 ■ (7v - 6w) 17 14. 4 6m + 5 • 4 m ~ 5 

10. (8a + 7c) 3 - (8a + 7c) 19 15. a; 8 ■ x 3 

11. 3 4 - 3° 16. a 9 -a 15 

12. 5 7 - 5° 17. 2 5 -2 3 

13. 4" • 4 8 "+ 3 18. 2 10 ■ 2 3 

In Exercises 19-28, simplify each of the given exponential expressions. 

4 16 nA (46 + 7c) 15 

19 416 



3 12 
20. — 

3 12 



21.— 

to' 

c io 
22. — r 



(9a -8c) 15 
' (9a-8c)8 





(46 + 7c) 5 




29n+5 


25. 


23n-4 




24fc-9 


26. 


23fe-8 


27. 


4 17 
49 


28. 


2 17 
2 6 



29. 


(4 8m ~ 


6 )' 


30. 


(2 2m " 


9\ 3 


31. 


[(to + 52/) 3 ] 7 


32. 


[(4u- 


-) 8 ] 9 


33. 


(43) 2 





354 CHAPTER 5. POLYNOMIAL FUNCTIONS 

In Exercises 29-38, simplify each of the given exponential expressions. 

34. (3 4 ) 2 

35. ( c y 

36. (w 9 ) 5 

37. (6 2 )° 

38. (8 9 )° 

In Exercises 39-48, simplify each of the given exponential expressions. 

39. (uw) 5 44. (-3w 9 ) 4 

40. {ac) 4 45. (-3x s y 2 ) 4 

41. (-2y) 3 46. (2x 8 z 6 ) 4 

42. {-2b) 3 47. (7s 6 ™) 3 

43. (3w 9 ) 4 48. (96 6 ") 3 

In Exercises 49-56, simplify each of the given exponential expressions. 

49. (- 

V2 

. 2^ 2 
51. 

u / 

q \ 3 

52. I-- 



53. 


(4) 4 


54. 


(-£)" 


55. 


(I)* 


56. 


\u 12 



5.5. LAWS OF EXPONENTS 



355 



57. Complete each of the laws of exponents presented in the following box, then use the results to 
simplify the expressions to the right of the box. 



a m a n 




? 


a m 




? 


a n 






(a m ) n 


= 


? 


(ab) m 


= 


? 


(!)" 


= 


7 



a 3 a 5 



,5\7 



_ = ? 
2 

(« b ) 7 = ? 

(a&) 9 = ? 
«*» ? 



m- j*- j*- Answers •*$ •** •** 



1. -64 
3. 1 

9 
7. 1 

9. (7w - 6w) 35 

11. 3 4 

13. 4 9n+3 

15. a; 11 

17. 2 8 

19. 1 

21. w 4 

23. (9a -8c) 7 

25. 2 6n+9 

27. 4 8 

OQ /|56m— 42 

31. (9s + 5y) 21 



33. 4 6 

35. c 28 
37. 1 
39. u 5 w 5 
41. -8y 3 
43. 81w 36 
45. 81x 32 y s 
47. 343s 18 " 



v 
49. — 



51. 



53. 



55. 



57. 



625 
625 

c 36 

25 



» 



21 



356 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Answer: -18s 4 i 6 



5.6 Multiplying Polynomials 

In this section we will find the products of polynomial expressions and func- 
tions. We start with the product of two monomials, then graduate to the 
product of a monomial and polynomial, and complete the study by finding the 
product of any two polynomials. 

The Product of Monomials 

As long as all of the operations are multiplication, we can use the commutative 
and associative properties to change the order and regroup. 



You Try It! 



Multiply: 3 (Ax) 



EXAMPLE 1. Multiply: -5(7j/). Simplify: -5(7y). 

Solution: Use the commutative and associative properties to change the order 
and regroup. 



-5(7y) 



[(-5)(7)]y 
-35y 



Reorder. Regroup. 
Multiply: (-5) (7) : 



-35. 



Answer: 12a; 



□ 



You Try It! 



Multiply: (7y 5 )(-2y 2 



EXAMPLE 2. Multiply: (-3a; 2 ) (4a; 3 ). Simplify: (-3a; 2 ) (4a; 3 ). 

Solution: Use the commutative and associative properties to change the order 
and regroup. 



(-3a; 2 )(4a; 3 ) = [(-3)(4)](xV) Reorder. Regroup. 
= -12a; 5 Multiply: (-3) (4) 



12, a; 2 x 3 =x 5 . 



Answer: 



-Uy 7 



□ 



You Try It! 



Multiply: (-6st 2 )(3s 3 t 



3+4\ 



EXAMPLE 3. Multiply: (-2a 2 6 3 )(-5a 3 6). Simplify: (-2a 2 6 3 )(-5a 3 6). 

Solution: Use the commutative and associative properties to change the order 
and regroup. 



(-2a 2 6 3 )(-5a 3 6) = [(-2)(-5)](aV)(6 3 6) Reorder. Regroup 



10a 5 6 4 



Multiply: (-2)(-5) = 10, 



a 2 a 3 



a b , and b A b = 6 4 



□ 



5.6. MULTIPLYING POLYNOMIALS 



357 



When multiplying monomials, it is much more efficient to make the required 
calculations mentally. In the case of Example 1, multiply —5 and 7 mentally 
to get 

-5(7y) = -35y. 

In the case of Example 2, multiply —3 and 4 to get —12, then repeat the base 
x and add exponents to get 



(-3a; 2 )(4a; 3 



-12a; a . 



Finally, in the case of Example 3, multiply —2 and —5 to get 10, then repeat 
the bases and add their exponents. 



■i;,4 



{-2a% 6 ){-ha 6 b) = 10a b 6 



Multiplying a Monomial and a Polynomial 

Now we turn our attention to the product of a monomial and a polynomial. 



EXAMPLE 4. Multiply: 5(3a: 2 - Ax - 8) 

Solution: We need to first distribute the 5 times each term of the polynomial. 
Then we multiply the resulting monomials mentally. 

5(3x 2 - 4a; - 8) = 5(3a; 2 ) - 5(4ar) - 5(8) 
= 15a; 2 - 20a; - 40 



You Try It! 



Multiply: A(-2y 2 + 3y + 5) 



Answer: -8y 2 + 12y + 20 



□ 



EXAMPLE 5. Multiply: 2j/(-3y + 5) 

Solution: We need to first distribute the 2y times each term of the polynomial. 
Then we multiply the resulting monomials mentally. 

2y(-3y + 5) = 2y(-3y) + 2y(5) 
= -6y 2 + lOy 



You Try It! 



Multiply: -3a: (4 - 2x) 



Answer: — 12a; + 6a; 2 



□ 



358 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Multiply: 

2xy(x 2 - 4xy 2 + 7x) 



EXAMPLE 6. Multiply: -3a6(a 2 + 2ab - b 2 ) 

Solution: We need to first distribute the — 3ab times each term of the poly- 
nomial. Then we multiply the resulting monomials mentally. 



-3ab(a 2 + 2ab - b 2 ) 



-3ab(a 2 ) + (-3ab)(2ab) - (-3ab){b 2 ) 
-3a 3 b + (-6a 2 b 2 ) - (Sab 3 ) 
-3a 3 b - 6a 2 b 2 + 3ab 3 



Answer: 

2x 3 y - 8x 2 y 3 + \Ax 2 y 



Alternate solution: It is far more efficient to perform most of the steps of 
the product — 3a6(a 2 + 2ab — b 2 ) mentally. We know we must multiply — 3ab 
times each of the terms of the polynomial a 2 + 2ab — b 2 . Here are our mental 
calculations: 

i) — 3ab times a 2 equals — 3a 3 b. 
ii) — 3ab times 2ab equals — 6a 2 b 2 . 
iii) — 3ab times — b 2 equals 3ab 3 . 

Thinking in this manner allows us to write down the answer without any 
of the steps shown in our first solution. That is, we immediately write: 



-3ab(a 2 + 2ab- b 2 



-3a 3 b-6a 2 b 2 + 3ab 3 



D 



You Try It! 



Multiply: -5y 3 {y 2 - 2y + 1 



EXAMPLE 7. Multiply: -2z 2 {z 3 + Az 2 - 11) 

Solution: We need to first distribute the — 2z 2 times each term of the poly- 
nomial. Then we multiply the resulting monomials mentally. 



-2z 2 {z 3 + 4z 2 



11) 



-2z 2 (z 3 ) + (-2z 2 ){Az 2 ) - (-2z 2 )(ll) 
-2z 5 + (-8z 4 )-(-22z 2 ) 



-2z b 



iz A + 22z 2 



Alternate solution: It is far more efficient to perform most of the steps of 
the product — 2z 2 (z 3 + 4z 2 — 11) mentally. We know we must multiply — 2z 2 
times each of the terms of the polynomial z 3 + 4z 2 — 11. Here are our mental 
calculations: 



5.6. MULTIPLYING POLYNOMIALS 



359 



i) — 2z 2 times z 3 equals — 2z 5 . 
ii) — 2z 2 times 4z 2 equals — 8z 4 . 
iii) — 2z 2 times —11 equals 22z 2 . 

Thinking in this manner allows us to write down the answer without any 
of the steps shown in our first solution. That is, we immediately write: 

-2z 2 (z 3 + 4z 2 - 11) = -2z 5 - 8z 4 + 22z 2 



Answer: — by 5 + 10y 4 — 5y 3 



□ 



Multiplying Polynomials 

Now that we've learned how to take the product of two monomials or the 
product of a monomial and a polynomial, we can apply what we've learned to 
multiply two arbitrary polynomials. 

Before we begin with the examples, let's revisit the distributive property. 
We know that 

2- (3 + 4) = 2-3 + 2-4. 

Both sides are equal to 14. We're used to having the monomial factor on the 
left, but it can also appear on the right. 

(3 + 4) -2 = 3-2 + 4-2 

Again, both sides equal 14. So, whether the monomial appears on the left or 
right makes no difference; that is, a(b + c) = ab + ac and (b + c)a = ba + ca 
give the same result. In each case, you multiply a times both terms of b + c. 



EXAMPLE 8. Multiply: (2x + 5) (3a; + 2) 

Solution: Note that (2x + 5) (3a; + 2) has the form (b + c)a, where a is the 
binomial 3a; + 2. Because (b + c)a = ba + ca, we will multiply 3a; + 2 times both 
terms of 2a; + 5 in the following manner. 

(2a; + 5)(3a; + 2) = 2a;(3a; + 2) + 5(3a; + 2) 
Now we distribute monomials times polynomials, then combine like terms. 



You Try It! 



Multiply: (3.x + 4) (5a 



6x z 



ix + 15a; + 10 



Thus, (2a; + 5)(3a; + 2) 



= 6x 2 + 19a; + 10 
6a; 2 + 19a; + 10. 



Answer: 15a; 2 + 23a; + 4 



□ 



360 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Multiply: 

(z + 4)(z 2 + 3z + 9) 



EXAMPLE 9. Multiply: (x + 5) (a; 2 + 2x + 7) 

Solution: Note that (a; + 5) (a; 2 + 2x + 7) has the form (b + c)a, where a is the 
binomial 3a; + 2. Because (6 + c)« = ba + ca, we will multiply a; 2 + 2a; + 7 times 
both terms of x + 5 in the following manner. 

(a; + 5)(.t 2 + 2x + 7) = x(x 2 + 2a; + 7) + 5(a; 2 + 2a; + 7) 
Now we distribute monomials times polynomials, then combine like terms. 

= x 3 + 2x 2 + 7x + 5x 2 + Wx + 35 



Answer: z 3 + 7z 2 + 21 z + 36 



= x 3 + 7x 2 + YIx + 35 
Thus, (a; + 5)(.x 2 + 2a; + 7) = a; 3 + 7a; 2 + 17a; + 35. 



□ 



Speeding Things Up a Bit 

Let's rexamine Example 9 with the hope of unearthing a pattern that will allow 
multiplication of polynomials to proceed more quickly with less work. Note the 
first step of Example 9. 

(x + 5){x 2 + 2x + 7) = a;(a; 2 + 2a; + 7) + 5(a; 2 + 2a; + 7) 

Note that the first product on the right is the result of taking the product 
of the first term of x + 5 and x 2 + 2x + 7. Similarly, the second product on 
the right is the result of taking the product of the second term of x + 5 and 
a; 2 + 2a; + 7. Next, let's examine the result of the second step. 

(x + 5)(x 2 + 2x + 7) = x 3 + 2x 2 + 7x + 5x 2 + Wx + 35 



The first three ter ms on the right are t he result of multiplying x times a; 2 +2x+7. 

2x + 7) 



1 



{x + 5)(x' + 2x + 7) 



2x A 



•7ar+' 



The second set of three terms on the right are the result of multiplying 5 times 

a; 2 + 2a; + 7. I J I 1 

5)(a; 2 + 2a; + 7) = - 



(a; + 5)(a 



5a;^ 



10a; + 35 



These notes suggest an efficient shortcut. To multiply x + 5 times a; 2 + 2a; + 7, 

• multiply x times each term of x 2 + 2x + 7, then 

• multiply 5 times each term of a; 2 + 2a; + 7. 

• Combine like terms. 



5.6. MULTIPLYING POLYNOMIALS 



361 



This process would have the following appearance. 



(a; + 5)(a; 2 + 2x + 7)=x 3 + 2x 2 + 7x + 5a; 2 + Wx + 35 

17a; + 35 



x 3 + 7x 2 



You Try It! 



EXAMPLE 10. Use the shortcut technique described above to simplify Multiply: 

(3a; + 2){Ax 2 - x + 10) 
(2y-6)(3 2 / 2 + 4 2 /+ll) 

Solution: Multiply 2y times each term of 3j/ 2 + Ay + 11, then multiply —6 
times each term of 3j/ 2 + Ay + 11. Finally, combine like terms. 

(2y - 6){3y 2 + Ay + 11) = 6y 3 + 8y 2 + 22y - 18y 2 - 2Ay - 66 
= 6y 3 - 10y 2 -2y-66 

Answer: 

12z 3 + 5a; 2 + 28a; + 20 



D 



EXAMPLE 11. Use the shortcut technique to simplify (a + b) 2 . 
Solution: To simplify (a + b) 2 , we first must write a + b as a factor two times. 

{a + bf = (a + b)(a + b) 

Next, multiply a times both terms of a + b, multiply b times both terms of 
a + b, then combine like terms. 

= a 2 + ab + ba + b 2 
= a 2 + 2ab + b 2 

Note that ab = ba because multiplication is commutative, so ab + ba = 2ab. 



You Try It! 



Multiply: (3y - 2) 



Answer: 9y 2 — 12y + A 



□ 



EXAMPLE 12. Use the shortcut technique to simplify 

(x 2 +x + l)(x 2 - x-l). 



You Try It! 



Multiply: 

(a 2 -2a + 3)(a 2 + 2a-3) 



362 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Solution: This time the first factor contains three terms, x 2 , x, and 1, so 



we first multiply x times each term of x 
x 2 — x — 1, and 1 times each term of x 2 — 

(x 2 + x + \){x 2 - x - 1) = x 4 - x 3 - 



1, then x times each term of 



x - 


1 


Then we 


combine like terms. 


x 2 


+ 


x 3 - x 2 - 


x + x 2 — x — 1 


2x 


— 


1 





Answer: a 4 - 4a 2 + 12a - 9 



a 



Some Applications 

Recall that the area of a circle of radius r is found using the formula A = irr 2 . 
The circumference (distance around) of a circle of radius r is found using the 
formula C = 2irr (see Figure 5.37). 




Figure 5.37: Formulae for the area and circumference of a circle of radius r. 



You Try It! 



Two concentric circles are 
shown below. The inner 
circle has radius x and the 
outer circle has radius x + 2. 
Find the area of the shaded 
region as a function of x. 




EXAMPLE 13. In Figure 5.38 are pictured two concentric circles (same 
center). The inner circle has radius x and the outer circle has radius x + I. 
Find the area of the shaded region (called an annulus) as a function of x. 




Figure 5.38: Find the area of the shaded region. 



5.6. MULTIPLYING POLYNOMIALS 



363 



Solution: We can find the area of the shaded region by subtracting the area 
of the inner circle from the area of the outer circle. 

A = Area of outer circle — Area of inner circle 

We use the formula A = irr 2 to compute the area of each circle. Because the 
outer circle has radius x + 1, it has area w(x + l) 2 . Because the inner circle has 
radius x, it has area irx 2 . 

= ir(x + 1) — irx 

Next, we'll expand (x + l) 2 , then combine like terms. 

= ir(x + l)(x + 1) — irx 
= n(x 2 + x + x + 1) — ttx 2 
= ir(x 2 + 2x + 1) - ttx 2 

Finally, distribute tt times each term of x + 2x + 1, then combine like terms. 

= TTX + 2lTX + TV — TTX 
= 2wX + TT 

Hence, the area of the shaded region is A = 2ttx + tt. 



Answer: Anx + An 



□ 



You Try It! 



EXAMPLE 14. The demand for widgets is a function of the unit price, 
where the demand is the number of widgets the public will buy and the unit 
price is the amount charged for a single widget. Suppose that the demand is 
given by the function x = 270 — 0.75p, where x is the demand and p is the unit 
price. Note how the demand decreases as the unit price goes up (makes sense). 
Use the graphing calculator to determine the unit price a retailer should charge 
for widgets so that his revenue from sales equals $20,000. 

Solution: To determine the revenue (R), you multiply the number of widgets 
sold (x) by the unit price (p). 

R = xp (5.5) 

However, we know that the number of units sold is the demand, given by the 
formula 

x = 270-0.75p (5.6) 

Substitute equation 5.6 into equation 5.5 to obtain the revenue as a function 
of the unit price. 

R = (270 - 0.75p)p 



Suppose that the demand for 
gadgets is given by the 
function x = 320 — 1.5p, 
where x is the demand and p 
is the unit price. Use the 
graphing calculator to 
determine the unit price a 
retailer should charge for 
gadgets so that her revenue 
from sales equals $12,000. 



364 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Expand. 



In this example, the 
horizontal axis is actually 
the p-axis. So when we set 
Xmin and Xmax, we're 
actually setting bounds on 
the p-axis. 



R = 270p - 0.75p 2 



(5.7) 



We're asked to determine the unit price that brings in a revenue of $20,000. 
Substitute $20,000 for R in equation 5.7. 



20000 = 270p - 0.75]/ 



(5.8) 



Enter each side of equation 5.8 into the Y= menu of your calculator (see 
the first image in Figure 5.39). After some experimentation, we settled on the 
WINDOW parameters shown in the second image of Figure 5.39. After making 
these settings, push the GRAPH button to produce the graph shown in the 
third image in Figure 5.39. 



Plotl PlotE Plots 
WiB270*X-0.75*X 
'-2 

W £020000 

sVs = 



WINDOW 
Xnin=0 
Xnax=400 
Xscl=100 
Ymin=0 
Vnax=30000 

Y£Cl=10000 

■i-Xres=l 




Figure 5.39: Solving 20000 = 270p- 0.75p 2 . 

Note that the third image in Figure 5.39 shows that there are two solutions, 
that is, two ways we can set the unit price to obtain a revenue of $20,000. To 
find the first solution, select 5:intersect from the CALC menu, press ENTER 
in response to "First curve," press ENTER in response to "Second curve," 
then move your cursor closer to the point of intersection on the left and press 
ENTER in response to "Guess." The result is shown in the first image in 
Figure 5.40. 

Repeat the process to find the second point of intersection. The result is 
shown in the second image in Figure 5.40. 




Intersection 
K=i04.£Bl££ ..Y=£0000. 



Intersection 
K=2££.7iB7H ..Y=£0000 . 



Figure 5.40: Finding the points of intersection. 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with p and R, respectively (see 
Figure 5.41). Include the units (dollars and dollars). 



Place your WINDOW parameters at the end of each axis (see Figure 5.41). 



5.6. MULTIPLYING POLYNOMIALS 



365 



30000 



, _R(dollars) 




R = 20000 



R = 270p - 0.75p 2 
p(dollars) 



104.28122 255.71878 400 

Figure 5.41: Reporting your graphical solution on your homework. 



• Label each graph with its equation (see Figure 5.41). 

• Drop a dashed vertical line through each point of intersection. Shade and 
label the p- values of the points where the dashed vertical lines cross the 
p-axis. These are the solutions of the equation 20000 = 270p — 0.75p 2 
(see Figure 5.41). 

Rounding to the nearest penny, setting the unit price at either $104.28 or 

$255.72 will bring in a revenue of $20,000. Answer: 



.55 or $164.79 



□ 



366 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



;».;-». t± 



Exercises 



•*;■*; -*i 



In Exercises 1-10, simplify the given expression. 

1. -3(7r) 

2. 7(3a) 

3. (-96 3 )(-86 6 ) 

4. (8s 3 )(-7s 4 ) 

5. (-7r 2 £ 4 )(7r 5 £ 2 ) 



6. 


(-10s 2 t 8 )(-7s 4 i 3 


7. 


(-56 2 c 9 )(-86 4 c 4 ) 


8. 


(-9s 2 i 8 )(7s 5 t 4 ) 


9. 


{-8v 3 )(4v 4 ) 


0. 


(-9y 3 )(3y 5 ) 



In Exercises 11-22, use the distributive property to expand the given expression. 



11. 9(-26 2 + 26 + 9) 

12. 9(-46 2 + 76-8) 

13. -4(10i 2 -7i-6) 

14. -5(-7w 2 -7u + 2) 

15. -8u 2 (-7u 3 - 8m 2 -2u + 10) 

16. -3s 2 (-7s 3 - 9s 2 + 6s + 3) 



17. 10s 2 (-10s 3 + 2s 2 + 2s + 8 

18. 8w 2 (9u 3 -5u 2 -2u + 5) 

19. 2st(-4s 2 + 8st - lQt 2 ) 

20. 7uv(-9u 2 - 3uv + 4v 2 ) 

21. -2uu;(10w 2 - 7uw - 2w 2 ) 

22. -6vw(-5v 2 + 9vw + 5w 2 ) 



In Exercises 23-30, use the technique demonstrated in Examples 8 and 9 to expand each of the following 
expressions using the distributive property. 



23. (-9x-4)(-3x + 2) 

24. (4x- 10)(-2a:-6) 

25. (3a; + 8)(3a;-2) 

26. (-6x+8)(-x+l) 



27. (2x- l)(-6x 2 +4x + 5) 

28. (Ax - 6)(-7ir 2 - lOcc + 10) 

29. (x-6)(-2x 2 -Ax -4) 

30. (5x- 10)(-3a; 2 + 7a;-8) 



In Exercises 31-50, use the shortcut technique demonstrated in Examples 10, 11, and 12 to expand 
each of the following expressions using the distributive property. 



31. (8u - 9w)(8u - 9w) 

32. (36 + 4c)(-86+10c) 



33. (9r- 7t)(3r-9t) 

34. (-6x-3y)(-6x + 9y) 



5.6. MULTIPLYING POLYNOMIALS 



367 



35. {At - 10s)(-10r 2 + lOrs - 7s 2 ) 

36. (5s - 9f)(-3s 2 + Ast - 9t 2 ) 

37. (9a; - 2z)(4x 2 - Axz - Wz 2 ) 

38. (r - 4i)(7r 2 + Art - 2t 2 ) 

39. (9r + 3i) 2 

40. (4a; + 8z) 2 

41. (Ay + 5z)(Ay - 5z) 

42. (7w + 2w)(7w-2w) 



43. (7u + 8v)(7u - 8v) 

44. (66 + 8c)(66-8c) 

45. (76 + 8c) 2 

46. (26 + 9c) 2 

47. (2i 2 + 9t + A)(2t 2 + 9t + A) 

48. (3a 2 - 9a + 4)(3a 2 - 9a + 2) 

49. (Aw 2 + 3w + 5)(3w 2 - 6w + 8) 

50. (4s 2 + 3s + 8)(2s 2 + 4s - 9) 



51. The demand for widgets is given by the 
function x = 320 — 0.95p, where x is the 
demand and p is the unit price. What unit 
price should a retailer charge for widgets 
in order that his revenue from sales equals 
$7,804 ? Round your answers to the near- 
est cent. 



52. The demand for widgets is given by the 
function x = 289 — 0.91p, where x is the 
demand and p is the unit price. What unit 
price should a retailer charge for widgets 
in order that his revenue from sales equals 
$7,257 ? Round your answers to the near- 
est cent. 



53. In the image that follows, the edge of 
the outer square is 6 inches longer than 
3 times the edge of the inner square. 



54. In the image that follows, the edge of 
the outer square is 3 inches longer than 
2 times the edge of the inner square. 



□ 




Express the area of the shaded region as a 
polynomial in terms of x, the edge of the 
inner square. Your final answer must be 
presented as a second degree polynomial 



in the form A(x) 



bx - 



b) Given that the edge of the inner square is 
5 inches, use the polynomial in part (a) to 
determine the area of the shaded region. 



Express the area of the shaded region as a 
polynomial in terms of x, the edge of the 
inner square. Your final answer must be 
presented as a second degree polynomial 



in the form A(x) 



bx + c. 



b) Given that the edge of the inner square is 
4 inches, use the polynomial in part (a) to 
determine the area of the shaded region. 



368 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



55. A rectangular garden is surrounded by a 
uniform border of lawn measuring x units 
wide. The entire rectangular plot mea- 
sures 31 by 29 feet. 



29 









X 















31 



Find the area of the interior rectangu- 
lar garden as a polynomial in terms of 
x. Your final answer must be presented 
as a second degree polynomial in the form 



A(x) 



bx + c. 



b) Given that the width of the border is 9.3 
feet, use the polynomial in part (a) to de- 
termine the area of the interior rectangular 
garden. 



56. A rectangular garden is surrounded by a 
uniform border of lawn measuring x units 
wide. The entire rectangular plot mea- 
sures 35 by 24 feet. 



24 









X 















35 



a) Find the area of the interior rectangu- 
lar garden as a polynomial in terms of 
x. Your final answer must be presented 
as a second degree polynomial in the form 
A(x) = ax 2 + bx + c. 

b) Given that the width of the border is 1.5 
feet, use the polynomial in part (a) to de- 
termine the area of the interior rectangular 
garden. 



j*- j*- j*- Answers •*$ **s ■** 



1. -21r 
3. 72b 9 

5. -49r 7 * 6 

7. 406 6 c 13 

9. -32m 7 
11. -186 2 + 186 + 81 
13. -40* 2 + 28* + 24 



15. 56m 5 + 64m 4 + 16m 3 - 80m 2 



17. -100s 5 + 20s 4 + 20s 3 + 80s 2 
19. -8s 3 * + 16s 2 * 2 - 20s* 3 



21. -20u 3 w + 14u 2 w 2 + 4mm; 3 

23. 27x 2 - 6a; - 8 

25. 9x 2 + 18a;- 16 

27. -12a; 3 + 14x 2 + 6a; - 5 

29. -2a; 3 + 8a; 2 + 20a; + 24 

31. 64m 2 - IUuw + 81w 2 

33. 27r 2 - 102r* + 63t 2 



35. -40r 3 + 140r 2 s - 128rs 2 + 70s 3 



37. 36a; 3 - 44a; 2 z - 82a;z 2 + 20z 3 
39. 81r 2 + 54r* + 9i 2 



5.6. MULTIPLYING POLYNOMIALS 



369 



41. 16y 2 - 25z 2 

43. 49w 2 - 64v 2 

45. 496 2 + 1126c + 64c 2 

47. 4f* + 36i 3 + 97£ 2 + 72t + 16 



49. I2w 4 - 15w 3 + 29w 2 - 6w + 40 

51. $26.47, $310.37 

53. AO) = 8a; 2 + 36x + 36, A(5) = 416 square 
inches 

55. 899 - 120a; + 4a; 2 , 128.96 square feet 



370 CHAPTER 5. POLYNOMIAL FUNCTIONS 

5.7 Special Products 

This section is dedicated to explaining a number of important shortcuts for mul- 
tiplying binomials. These are extremely important patterns that will produce 
the same products computed in previous sections. It is essential that readers 
practice until they become proficient using each of the patterns presented in 
this section. 

The FOIL Method 

Consider the product of two binomials (a; + 3)(a; + 6). We already know how 
to find the product of these two binomials; we multiply x times both terms of 
x + 6, then we multiply 3 times both terms of x + 6. 

(x + 3)(x + 6) = x 2 + 6x + 3a; + 18 

Normally we combine like terms, but we halt the process at this point so as to 
introduce the pattern called the FOIL method. 

The letters in the word FOIL stand for "First," "Outer," "Inner," and 
"Last." Let's see how we can connect these terms to the product (a; + 3) (a; + 6). 

• The arrows indicate the terms in the "First" positions in each binomial. 
If you multiply the terms in the "First" position, you get x 2 . 

( |+3)(| + 6) 
F 

• The arrows indicate the terms in the "Outer" positions in each binomial. 
If you multiply the terms in the "Outer" positions, you get 6x. 

( f + 3)(s + f ) 
O 

• The arrows indicate the terms in the "Inner" positions in each binomial. 
If you multiply the terms in the "Inner" positions, you get 3x. 

(i + JKj + 6) 
I 

• The arrows indicate the terms in the "Last" positions in each binomial. 
If you multiply the terms in the "Last" positions, you get 18. 

(E + ^QE + g ) 

L 



5.7. SPECIAL PRODUCTS 



371 



The following diagram shows the connection between "First," "Outer," "In- 
ner," "Last," and the answer. 

FOIL 

(x + 3){x + 6) = x 2 + 6x + 3a; + 18 



EXAMPLE 1. Use the FOIL method to simplify: (x + 5) (a; + 7) 

Solution: Multiply the "First" positions: x 2 . Multiply the "Outer" positions: 
7x. Multiply the "Inner" positions: 5a;. Multiply the "Last" positions: 35. 



F 

„2 



o 

7x 



(x + 5)(x + 7) -- 
Combining like terms, {x + 5) (a; + 7) = x 2 + 12a; + 35 



I 

5;); 



L 
35 



You Try It! 



Simplify: (x + 2)(x + 11) 



Answer: 



13a: + 22 



D 



EXAMPLE 2. Use the FOIL method to simplify: (2x - 7) (x - 4) 

Solution: Multiply the "First" positions: 2a; 2 . Multiply the "Outer" posi- 
tions: —8a;. Multiply the "Inner" positions: —7a;. Multiply the "Last" posi- 
tions: 28. 



You Try It! 



Simplify: (a;- l)(4a; + 5) 



(2a; -7)0 -4) 
Combining like terms, (2a; — 7) (a; — 4) = 2a; 



/•' 


O I 


L 


2x 2 


- 8a; - 7a; + 


28 


-4) = 


2a; 2 - 15a; + 28. 





Answer: 4a; 2 + x — 5 



□ 



At first glance, the FOIL method doesn't look like much of a shortcut. After 
all, if we simply use the distributive property on the product of Example 2, we 
get the same quick result. 

(2a; - 7)(x - 4) = 2a;(a; - 4) - 7{x - 4) 
= 2a; 2 - 8a; - 7a; + 28 
= 2a; 2 - 15a; + 28 

The FOIL method becomes a true shortcut when we add the "Outer" and 
"Inner" results in our head. 



372 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



FOIL Shortcut. To multiply two binomials, follow these steps: 

1. Multiply the terms in the "First" positions. 

2. Multiply the terms in the "Outer" and "Inner" positions and combine 
the results mentally (if they are like terms) . 

3. Multiply the terms in the "Last" positions. 



Simplify: (2z - 3)(5z - 1) EXAMPLE 3. Use the FOIL shortcut to simplify: (3x + 8) (2a - 1) 

Solution: Each of the following steps is performed mentally. 

1. Multiply the terms in the "First" positions: 6x 2 

2. Multiply the terms in the "Outer" and "Inner" positions and add the 
results mentally: — 3.T + 16a; = 13a; 

3. Multiply the terms in the "Last" positions: —8 

Write the answer with no intermediate steps: (3a; + 8) (2a; — 1) = 6a; 2 + 13a; — 8. 
Answer: 10z 2 - 17z + 3 

□ 



You Try It! 



Simplify: (7a; + 2) (2a; - 3) 



Answer: 14a; 2 — 17a; — 6 



EXAMPLE 4. Use the FOIL shortcut to simplify: (Ay - 3)(5y + 2) 
Solution: Each of the following steps is performed mentally. 

1. Multiply the terms in the "First" positions: 20y 2 

2. Multiply the terms in the "Outer" and "Inner" positions and add the 
results mentally: 8y — 15y = —7y 

3. Multiply the terms in the "Last" positions: —6 

Write the answer with no intermediate steps: (Ay — 3)(5y + 2) = 20j/ 2 — 7y — 6. 

□ 



5.7. SPECIAL PRODUCTS 373 

The Difference of Squares 

We can use the FOIL shortcut to multiply (a + 6) (a — b). 

1. Multiply the terms in the "First" positions: a 2 

2. Multiply the terms in the "Outer" and "Inner" positions and add the 
results mentally: ab — ab = 

3. Multiply the terms in the "Last" positions: — b 2 

Thus, (a + b)(a — b) = a 2 — b 2 . Note how the right-hand side a 2 — b 2 is the 
difference of two squares. This leads to the following shortcut. 



The 


difference 


of squares. 


If you have identical terms in 


the 


'First" po- 


sitions and identical terms in 


the "Last" positions, 


but one 


set is 


separated 


with 


a plus sign l 


while the other is separated by a minus sign, 


then 


proceed as 


follows: 












1. 


Square the 


"First" term 










2. 


Square the 


"Last" term. 










3. 


Place a minus sign between the results 








That 


is, 
















(a- 


f b)(a-b) = a 2 -b 2 









Note: If you don't have identical terms in the "First" and "Last" positions, 
with one set separated with a plus sign and the other with a minus sign, then 
you do not have the difference of squares pattern and you must find some other 
way to multiply. For example, (x + 3)(x — 3) is an example of the difference of 
squares pattern, but (2y + 3)(2y — 5) is not. 



You Try It! 



EXAMPLE 5. Use the difference of squares shortcut to simplify: (x + 3)(x — Simplify: (x + 5)(x — 5) 
3) 

Solution: Note how the terms in the "First" position are identical, as are the 
terms in the "Last" position, with one set separated by a plus sign and the 
other with a minus sign. Hence, this is the difference of squares pattern and 
we proceed as follows: 

1. Square the term in the "First" position: x 2 

2. Square the term in the "Last" position: (— 3) 2 = 9 

3. Separate the squares with a minus sign. 



374 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Answer: x 2 — 25 



That is: 



(a; + 3)(x-3) = x 2 - (3) 2 



9 



Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 



as in (x + 3)(x — 3) 



9. 



□ 



You Try It! 



Simplify: (3a - 66) (3a + 66) 



Answer: 9a 2 — 366 2 



EXAMPLE 6. Use the difference of squares shortcut to simplify: 

(8y + 7z)(8y-7z) 

Solution: Note how the terms in the "First" position are identical, as are the 
terms in the "Last" position, with one set separated by a plus sign and the 
other with a minus sign. Hence, this is the difference of squares pattern and 
we proceed as follows: 

1. Square the term in the "First" position: (8y) 2 = 64y 2 

2. Square the term in the "Last" position: (7z) 2 = 49z 2 

3. Separate the squares with a minus sign. 
That is: 

(8y + 7z)(8y-7z) = (8y) 2 -(7z) 2 
= 64y 2 - 49z 2 

Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 
as in (8y + 7z)(8y - 7z) = 64y 2 - AQz 2 . 

□ 



You Try It! 



Simplify: 

(2y A + z 3 )(2y A - z 3 ) 



EXAMPLE 7. Use the difference of squares shortcut to simplify: 

(x 3 - 5y 2 )(x 3 + by 2 ) 

Solution: Note how the terms in the "First" position are identical, as are the 
terms in the "Last" position, with one set separated by a plus sign and the 
other with a minus sign. Hence, this is the difference of squares pattern and 
we proceed as follows: 



5.7. SPECIAL PRODUCTS 375 

1. Square the term in the "First" position: (a; 3 ) 2 = x 6 

2. Square the term in the "Last" position: (5y 2 ) 2 = 25y 4 

3. Separate the squares with a minus sign. 
That is: 

(a; 3_ 5y 2 )(a; 3 +52/ 2 ) = (a; 3 ) 2_ (% 2 ) 2 

= x 6 - 25y 4 

Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 
as in (x 3 - 5y 2 )(x 3 + by 2 ) = x 6 - 25y 4 . Answer: 4y 8 - z 6 



Squaring a Binomial 

Before demonstrating the correct procedure for squaring a binomial, we will 
first share one of the most common mistakes made in algebra. 



□ 



Warning! This is incorrect! 


One 


of the most 


common mistakes made in 


algebra is the assumption 


that: 












(a 


+ bf 


= a 2 + b 2 






The fact that this is incorrect is 


easily 


checked. Substitute 3 for 


a and 4 for b. 




(3 


+ 4) 2 

? 2 


= 3 2 + 4 2 
= 3 2 + 4 2 










49 


= 9 + 16 






Clearly this is incorrect! 













So what is the correct answer? First, (a + b) 2 = (a + b)(a + b). We can now 
use the FOIL shortcut. 

1. Multiply the terms in the "First" positions: a 2 

2. Multiply the terms in the "Outer" and "Inner" positions and add the 
results mentally: ab + ab = 2ab 

3. Multiply the terms in the "Last" positions: b 2 

Hence, the correct answer is (a + b) 2 = a 2 + 2ab + b 2 . This leads us to the 
following shortcut for squaring a binomial. 



376 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



Squaring a binomial. 

following steps: 


To square a 


binomial, such as (a + 


b) 2 , 


perform 


the 


1. 


Square the 


"First" 


term: 


a 2 










2. 


Multiply the "First" and 


"Last' 


terms and double the 


result: 2ab 




3. 


Square the 


"Last" 


term: 


b 2 










That 


is: 




(o4 


bf = 


a 2 + 2ab + b 2 









You Try It! 



Simplify: (a; + 3) 



Answer: x 2 + 6x + 9 



EXAMPLE 8. Use the squaring a binomial shortcut to expand: (x + 5) 2 
Solution: Follow these steps: 

1. Square the first term: x 2 

2. Multiply the "First" and "Last" terms and double the result: 2(x)(5) = 
10a; 

3. Square the "Last" term: 5 2 = 25 

Thus: 

(a; + 5) 2 =x: 2 + 2(x:)(5) + (5) 2 
= x 2 + Wx + 25 

Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 
as in (x + 5) 2 = x 2 + Wx + 25. 

□ 



Comment. Students often refuse to learn the "squaring a binomial" shortcut, 
noting that they can just as easily use the FOIL technique or a simple appli- 
cation of the distributive property to arrive at the same result. Unfortunately, 
failure to learn the "squaring a binomial" shortcut will severely handicap stu- 
dents, as this pattern is an important component of many procedures in future 
mathematics courses. 



You Try It! 



Simplify: (2y + 3z) 2 



EXAMPLE 9. Use the squaring a binomial shortcut to expand: (3x -I- 7y) 2 
Solution: Follow these steps: 



5.7. SPECIAL PRODUCTS 377 

1. Square the first term: (3a:) 2 = 9x 2 

2. Multiply the "First" and "Last" terms and double the result: 2(3x)(7y) = 
A2xy 

3. Square the "Last" term: (7y) 2 = 49j/ 2 
Thus: 

(3i + 7y) 2 = (3a;) 2 + 2(3a:)(7y) + (7y) 2 
= 9a; 2 + 42a;?/ + 49j/ 2 

Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 
as in (3a; + 7y) 2 = 9.x 2 + A2xy + 49y 2 . Answer: Ay 2 + 12yz + 9z 2 



In the next example, when squaring a binomial with a minus sign, we take 
care of the minus sign by "adding the opposite." 



You Try It! 



EXAMPLE 10. Use the squaring a binomial shortcut to expand: (4a 2 — 5o 3 ) 2 Simplify: (3a; 4 — 5z 2 ) 2 

Solution: Add the opposite: (4a 2 -56 3 ) 2 = (4a 2 + (-56 3 )) 2 . Now follow these 
steps: 

1. Square the first term: (4a 2 ) 2 = 16a 4 

2. Multiply the "First" and "Last" terms and double the result: 2(4a 2 )(-5o 3 ) = 
-40a 2 6 3 

3. Square the "Last" term: (-56 3 ) 2 = 25b 6 
Thus: 

(4a 2 - 56 3 ) 2 = (4a 2 + (-56 3 )) 2 

= (4a 2 ) 2 + 2(4a 2 )(-56 3 ) + (-56 3 ) 2 
= 16a 4 - 40a 2 6 3 + 25o 6 

Note: You should practice this pattern until you can go straight from the 
problem statement to the answer without writing down any intermediate work, 
as in (4a 2 - 56 3 ) 2 = 16a 4 - 40a 2 6 3 + 25b 6 . Answer: 

9a; 8 - 30a; 4 ;z 2 + 25z 4 

Example 10 shows us that if we are squaring a difference, the middle term 
will be minus. That is, the only difference between (a + o) 2 and (a — 6) 2 is the 
sign of the middle term. 



a 



a 



378 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



You Try It! 



Simplify: (a 2 - 36 5 ) 5 



Answer: a 4 - 6a 2 6 5 + 96 10 



Squaring 


a 


binomial. 


The shortcuts for squaring a binomial 


are: 








{a + bf 


= a 2 + 2ab + b 2 












{a-bf 


= a 2 - 2ab + b 2 







EXAMPLE 11. Use the squaring a binomial shortcut to expand: (x 3 



y 



:?\2 



Solution: Use the pattern (a — b) z 



2at 



b 2 . Square the "First" term, 



multiply the "First" and "Last" terms and double the result, then square the 
"Last" term. Because of the minus sign, the middle term will be minus, but 
all other terms are plus. 



: '\2 



{x* - V y = {x s f - 2{x i ){y i ) + (p") 



2x 3 y 3 



D 



An Application 

In Example 7 on page 348, we found the area of the outer square by summing 
the areas of its parts (see Figure 5.42). Recall that the answer was A = x 2 + 
6x + 9. 





X 


3 


3 






x 







Figure 5.42: Find the area of the square. 



Now that we have the squaring a binomial shortcut, we can simplify the process 
of finding the area of the outer square by squaring its side. That is: 



5.7. SPECIAL PRODUCTS 379 



Now we can use the squaring the binomial technique to expand. 

= x 2 + 2(a;)(3) + (3) 2 
= x 2 + 6x + 9 

Note that this is the same as the answer found by summing the four parts of 
the square in Example 7. 



380 CHAPTER 5. POLYNOMIAL FUNCTIONS 



f> t* t* Exercises «•* -*$ •** 

In Exercises 1-12, use the FOIL shortcut as in Examples 3 and 4 to multiply the given binomials. 

1. (5a; + 2)(3x + 4) 7. (6a; - 2) (3a: - 5) 

2. (5a; + 2)(4x + 3) 8. (2x - 3) (6a; - 4) 

3. (6a; - 3)(5x + 4) 9. (6a; + 4)(3aj + 5) 

4. (6a; -2) (4x + 5) 10. (3a; + 2) (4a; + 6) 

5. (5a; -6) (3a; -4) 11. (4a; - 5) (6a; + 3) 

6. (6a; -4) (3a; -2) 12. (3a; - 5) (2a; + 6) 

In Exercises 13-20, use the difference of squares shortcut as in Example 5 to multiply the given bino- 
mials. 

13. (10a: - 12) (10a; + 12) 17. (3a; + 10) (3a; - 10) 

14. (10a; - 11) (10a; + 11) 18. (12a; + 12) (12a; - 12) 

15. (6a; + 9) (6a; - 9) 19. (10a; - 9)(10a; + 9) 

16. (9a; + 2) (9a; -2) 20. (4a; - 6) (4a; + 6) 

In Exercises 21-28, use the squaring a binomial shortcut as in Example 8 to expand the given expression. 

21. (2a; + 3) 2 25. (7a: + 2) 2 

22. (8a; + 9) 2 26. (4a; + 2) 2 

23. (9a; - 8) 2 27. (6a; - 5) 2 

24. (4a; - 5) 2 28. (4a: - 3) 2 

In Exercises 29-76, use the approrpiate shortcut to multiply the given binomials. 

29. (11a: - 2) (Use + 2) 33. (56 + 6c)(36 - 2c) 

30. (6a; - 7) (6a; + 7) 34. (3r + 2£)(5r - 3t) 

31. {7r-5tf 35. (3u + 5v)(3u - 5v) 

32. (11m - 9w) 2 36. (11a + 4c) (11a - 4c) 



5.7. SPECIAL PRODUCTS 



381 



37. (96 3 + 10c 5 )(96 3 - 10c 5 

38. (9r 5 + 7£ 2 )(9r 5 -7i 2 ) 

39. (9s - At)(9s + At) 

40. (12x - 7y)(12x + 7y) 

41. (7x - 9y)(7x + 9y) 

42. (lOr- llt)(10r+llt) 

43. (6a-66)(2a + 36) 

44. (6r - 5i)(2r + 3i) 

45. (10a;- 10)(10x + 10) 

46. (12a;-8)(12a: + 8) 

47. (4a + 26) (6a - 36) 

48. (36 + 6c)(26-4c) 

49. (56-4c)(36 + 2c) 

50. (36-2c)(46 + 5c) 

51. (46-6c)(66-2c) 

52. (Ay - Az)(5y - 3z) 

53. (llr 5 + 9£ 2 ) 2 

54. (llar 3 + 10;z 5 ) 2 

55. (Au - Av)(2u - 6v) 

56. (4m — 5w)(5u — 6w) 



57. 


8r 4 + 7t b ) 2 


58. 


[2x 5 + 5y 2 ) 2 


59. 


[Ar + 3t)(4r-3t) 


60. 


;3r + 4s)(3r-4s) 


61. 


5r + 6t) 2 


62. 


[12v + 5w) 2 


63. 


[3x - 4)(2x + 5) 


64. 


;5a;-6)(4a; + 2) 


65. 


;66 + 4c)(26 + 3c) 


66. 


'3v + 6w)(2v + Aw) 


67. 


;11m 2 + 8w 3 )(11u 2 -8w 3 


68. 


;3u 3 + llw 4 )(3u 3 - llw 4 


69. 


;4y + 3z) 2 


70. 


116 + 3c) 2 


71. 


[7u - 2v) 2 


72. 


[Ab - 5c) 2 


73. 


'3v + 2w)(5v + 6w) 


74. 


[by + 3z)(Ay + 2z) 


75. 


to-3)(6:c + 2) 


76. 


;6x-5)(3a; + 2) 



For each of the following figure, compute the area of the square using two methods, 
i) Find the area by summing the areas of its parts (see Example 7 on page 348). 
ii) Find the area by squaring the side of the square using the squaring a binomial shortcut. 



77. 



78. 





X 


10 









X 









X 


13 


13 






X 







382 



CHAPTER 5. POLYNOMIAL FUNCTIONS 



79. A square piece of cardboard measures 12 inches on each side. Four squares, each having a side 
of x inches, are cut and removed from each of the four corners of the square piece of cardboard. 
The sides are then folded up along the dashed lines to form a box with no top. 



b) 



12 





X X 




X 

X 




X 
X 




X X 







/ 


/ / 




V / 




/ 



12 

Find the volume of the box as a function of x, the measure of the side of each square cut from 
the four corner of the original piece of cardboard. Multiply to place your answer in standard 
polynomial form, simplifying your answer as much as possible. 

Use the resulting polynomial to determine the volume of the box if squares of length 1.25 inches 
are cut from each corner of the original piece of cardboard. Round your answer to the nearest 
cubic inch. 



80. Consider again the box formed in Exercise 79. 

a) Find the surface area of the box as a function of x, the measure of the side of each square cut 
from the four corner of the original piece of cardboard. Multiply to place your answer in standard 
polynomial form, simplifying your answer as much as possible. 

b) Use the resulting polynomial to determine the surface area of the box if squares of length 1.25 
inches are cut from each corner of the original piece of cardboard. Round your answer to the 
nearest square inch. 



£»• s*> £*■ Answers ■** •** ■*$ 



1. 15a; 2 + 26a; + 8 
3. 30a; 2 + 9a;- 12 
5. 15a; 2 - 38a; + 24 
7. 18a; 2 - 36a; + 10 
9. 18a; 2 + 42a; + 20 
11. 24a; 2 - 18a;- 15 



13. 100a; 2 - 144 
15. 36a; 2 - 81 
17. 9a; 2 - 100 
19. 100a; 2 - 81 
21. 4x 2 + 12a; + 9 
23. 81a; 2 - 144a; + 64 



5.7. SPECIAL PRODUCTS 



383 



25. 49a; 2 + 28a; + 4 

27. 36a; 2 - 60a; + 25 

29. 121a; 2 - 4 

31. 49r 2 - 70rt + 2bt 2 

33. 156 2 + 86c - 12c 2 

35. 9m 2 - 25v 2 

37. 816 6 - 100c 10 

39. 81s 2 - 16t 2 

41. 49a; 2 - 81y 2 

43. 12a 2 + 6a6 - 18b 2 

45. 100a; 2 - 100 

47. 24a 2 - 6b 2 

49. 156 2 - 26c - 8c 2 

51. 246 2 - 446c + 12c 2 

53. 121r 10 + 198r 5 t 2 + 81i 4 



55. 8m 2 - 32uv + 2Av 2 

57. 64r 8 + 112r 4 t 5 + 49i 10 

59. 16r 2 - 9t 2 

61. 25r 2 + 60rt + 36t 2 

63. 6a; 2 + 7a; - 20 

65. 126 2 + 266c + 12c 2 

67. 121w 4 - 64w 6 

69. 16y 2 + 24yz + 9z 2 

71. 49m 2 -28uv + 4v 2 

73. 15w 2 + 28vw + 12w 2 

75. 30a; 2 - 8a; - 6 

77. A = x 2 + 20x + 25 

79. a) V(x) = 144a;-48a; 2 +4a; 3 b) V{2) = 128 
cubic inches 



Chapter 6 



Factoring 



The ancient Babylonians left the earliest evidence of the use of quadratic equa- 
tions on clay tablets dating back to 1800 BC. They understood how the area 
of a square changes with the length of its side. For example, they knew it was 
possible to store nine times more bales of hay in a square loft if the side of the 
loft was tripled in length. However, they did not know how to calculate the 
length of the side of a square starting from a given area. The word "quadratic" 
comes from "quadratus," the Latin word for "square." In this chapter, we will 
learn how to solve certain quadratic equations by factoring polynomials. 



385 



386 



CHAPTER 6. FACTORING 



You Try It! 



6.1 The Greatest Common Factor 

We begin this section with definitions of factors and divisors. Because 24 = 
2 ■ 12, both 2 and 12 are factors of 24. However, note that 2 is also a divisor 
of 24, because when you divide 24 by 2 you get 12, with a remainder of zero. 
Similarly, 12 is also a divisor of 24, because when you divide 24 by 12 you get 
2, with a remainder of zero. 



Factors and divisors. Suppose m and n are integers. Then mis a divisor 
(factor) of n if and only if there exists another integer k so that n = m ■ k. 



The words divisor and factor are equivalent. They have the same meaning. 



List the positive divisors of EXAMPLE 1. List the positive divisors (factors) of 24. 

18. 

Solution: First, list all possible ways that we can express 24 as a product of 

two positive integers: 

24 = 1 ■ 24 or 24 = 2 ■ 12 or 24 = 3 ■ 8 or 24 = 4 ■ 6 

Answer: 1,2,3,6,9, and 18 Therefore, the positive divisors (factors) of 24 are 1, 2, 3, 4, 6, 8, and 24. 

□ 



You Try It! 



List the positive divisors that EXAMPLE 2. List the positive divisors (factors) that 36 and 48 have in 
40 and 60 have in common. common. 

Solution: First, list all positive divisors (factors) of 36 and 48 separately then 
box the divisors that are in common. 



12 



Divisors of 36 are: |T| , |J] , |T| , |T| , |T| , 9, 

Divisors of 48 are: [T| , [T| , [Y| , [T| , [T| , 8,|~12~|, 16, 24, 48 



Answer: 1,2,4,5,10, and 20 



Therefore, the common positive divisors (factors) of 36 and 48 are 1, 2, 3, 4, 
6, and 12. 

□ 



6. 1 . THE GREATEST COMMON FACTOR 



387 



Greatest common divisor. The greatest common divisor (factor) of a and b 
is the largest positive number that divides evenly (no remainder) both a and b. 
The greatest common divisor of a and 6 is denoted by the symbolism GCD(a, b). 
We will also use the abbreviation GCF(a, b) to represents the greatest common 
factor of a and b. 



Remember, greatest common divisor and greatest common factor have the 
same meaning. In Example 2, we listed the common positive divisors of 36 and 
48. The largest of these common divisors was 12. Hence, the greatest common 
divisor (factor) of 36 and 48 is 12, written GCD(36, 48) = 12. 

With smaller numbers, it is usually easy to identify the greatest common 
divisor (factor). 



You Try It! 



EXAMPLE 3. State the greatest common divisor (factor) of each of the State the greatest common 
following pairs of numbers: (a) 18 and 24, (b) 30 and 40, and (c) 16 and 24. divisor of 36 and 60. 

Solution: In each case, we must find the largest possible positive integer that 
divides evenly into both the given numbers. 

a) The largest positive integer that divides evenly into both 18 and 24 is 6. 
Thus, GCD(18,24) = 6. 

b) The largest positive integer that divides evenly into both 30 and 40 is 10. 
Thus, GCD(30,40) = 10. 

c) The largest positive integer that divides evenly into both 16 and 24 is 8. 
Thus, GCD(16,24) = 8. 

Answer: 12 



With larger numbers, it is harder to identify the greatest common divisor 
(factor). However, prime factorization will save the day! 



□ 



EXAMPLE 4. Find the greatest common divisor (factor) of 360 and 756. 
Solution: Prime factor 360 and 756, writing your answer in exponential form. 



You Try It! 



Find the greatest common 
divisor of 120 and 450. 



388 CHAPTER 6. FACTORING 



360 756 



36 10 9 

/ \ ($f \) (sf (D ci) fa 

3) 4 



Thus: 



360 = 2 3 • 3 2 ■ 5 
756 = 2 2 • 3 3 ■ 7 



indexgreatest common divisorlusing prime factorization To find the greatest 
common divisor (factor), list each factor that appears in common to the highest 
power that appears in common. 

In this case, the factors 2 and 3 appear in common, with 2 2 being the high- 
est power of 2 and 3 2 being the highest power of 3 that appear in common. 
Therefore, the greatest common divisor of 360 and 756 is: 

GCD(360,756) = 2 2 -3 2 
= 4-9 
= 36 

Therefore, the greatest common divisor (factor) is GCD(360,756) = 36. Note 
what happens when we write each of the given numbers as a product of the 
greatest common factor and a second factor: 

360 = 36-10 
756 = 36-21 

In each case, note how the second second factors (10 and 21) contain no addi- 
Answer: 30 tional common factors. 

□ 



Finding the Greatest Common Factor of Monomials 

Example 4 reveals the technique used to find the greatest common factor of 
two or more monomials. 



6. 1 . THE GREATEST COMMON FACTOR 



389 



Finding the GCF of two or more monomials. To find the greatest com- 
mon factor of two or more monomials, proceed as follows: 

1 . Find the greatest common factor (divisor) of the coefficients of the given 
monomials. Use prime factorization if necessary. 

2. List each variable that appears in common in the given monomials. 

3. Raise each variable that appears in common to the highest power that 
appears in common among the given monomials. 



EXAMPLE 5. Find the greatest common factor of 6x 3 y 3 and 9x 2 y 5 . 
Solution: To find the GCF of 6x 3 y 3 and 9x 2 y 5 , we note that: 

1. The greatest common factor (divisor) of 6 and 9 is 3. 

2. The monomials Qx 3 y 3 and 9x 2 y 5 have the variables x and y in common. 

3. The highest power of x in common is x 2 . The highest power of y in 
common is y 3 . 

Thus, the greatest common factor is GCF(6cc 3 j/ 3 , 9x 2 y 5 ) = 2>x 2 y 3 . Note what 
happens when we write each of the given monomials as a product of the greatest 
common factor and a second monomial: 



6x 3 y 3 



3x 2 y 3 ■ 2x 



9 X 2 y 5 = 3 X 2 y 3 ■ 3y 

Observe that the set of second monomial factors (2a; and 3j/) contain no addi- 
tional common factors. 



You Try It! 



Find the greatest common 
factor of 16xy 3 and 12x 4 y 2 . 



Answer: Axy 2 



□ 



EXAMPLE 6. Find the greatest common factor of 12.x 4 , 18a: 3 , and 30a; 2 . 
Solution: To find the GCF of 12a; 4 , 18a; 3 , and 30a; 2 , we note that: 

1. The greatest common factor (divisor) of 12, 18, and 30 is 6. 

2. The monomials 12a; 4 , 18a; 3 , and 30a; 2 have the variable x in common. 

3. The highest power of x in common is a; 2 . 



You Try It! 



Find the greatest common 
factor of 6y 3 , 15y 2 , and 9j/ 5 . 



390 



CHAPTER 6. FACTORING 



Thus, the greatest common factor is GCF(12a; 4 , 18a; 3 , 30a; 2 ) = 6a: 2 . Note what 
happens when we write each of the given monomials as a product of the greatest 
common factor and a second monomial: 



12a; 4 
18a; 3 
30a: 2 



6a; 


■2x 


6a; 


■ 2>x 


6x 2 


•5 



Answer: 3y 2 



Observe that the set of second monomial factors (2a; 2 , 3a;, and 5) contain no 
additional common factors. 

□ 



Factor Out the GCF 

In Chapter 5, we multiplied a monomial and polynomial by distributing the 
monomial times each term in the polynomial. 

2a;(3a; 2 + 4a; - 7) = 2x ■ 3x 2 + 2x ■ 4x - 2x ■ 7 
= 6x 3 + 8a; 2 - 14a; 

In this section we reverse that multiplication process. We present you with the 
final product and ask you to bring back the original multiplication problem. 
In the case 6a; 3 + 8a; 2 — 14a;, the greatest common factor of 6a; 3 , 8a; 2 , and 14a; 
is 2a;. We then use the distributive property to factor out 2a; from each term 
of the polynomial. 

6a; 3 + 8a; 2 - 14a; = 2x ■ 3a; 2 + 2a; • 4a; - 2x • 7 
= 2a;(3a; 2 + 4a; - 7) 



Factoring. Factoring is "unmultiplying." You are given the product, then 
asked to find the original multiplication problem. 



First rule of factoring. If the terms of the given polynomial have a greatest 
common factor (GCF), then factor out the GCF. 



Let's look at a few examples that factor out the GCF. 



You Try It! 



Factor: 9y 2 - I5y + 12 



EXAMPLE 7. Factor: 6a; 2 + 10a; + 14 



6. 1 . THE GREATEST COMMON FACTOR 



391 



Solution: The greatest common factor (GCF) of 6a; 2 , 10a: and 14 is 2. Factor 
out the GCF. 

6a; 2 + 10a; + 14 = 2 ■ 3a; 2 + 2 ■ 5x + 2 ■ 7 
= 2(3a; 2 + 5a; + 7) 

Checking your work. Every time you factor a polynomial, remultiply to 
check your work. 

Check: Multiply. Distribute the 2. 

2(3a; 2 + 5x + 7) = 2 ■ 3a; 2 + 2 ■ 5x + 2 ■ 7 
= 6a: 2 + 10a: + 14 

That's the original polynomial, so we factored correctly. 



Answer: 3(3y 2 — 5y + 4) 



□ 



You Try It! 



EXAMPLE 8. Factor: 12y 5 - 32y 4 + 8y 2 

Solution: The greatest common factor (GCF) of I2y 5 , 32j/ 4 and 8y 2 is Ay 2 . 
Factor out the GCF. 

12y 5 - 32y 4 + 8y 2 = 4y 2 • 3y 3 - Ay 2 ■ 8y 2 + Ay 2 ■ 2 
= 4y 2 (3j/ 3 -8 2 / 2 + 2) 

Check: Multiply. Distribute the monomial Ay 2 . 

Ay 2 {3y 3 - 8y 2 + 2) = 4y 2 • 3y 3 - Ay 2 ■ 8y 2 + Ay 2 ■ 2 
= 12y 5 - 32y 4 + 8y 2 

That's the original polynomial. We have factored correctly. 



Factor: 8a; 6 + 20.x 4 - 24a; 3 



Answer: 4a; 3 (2x 3 + 5x - 6) 



□ 



EXAMPLE 9. Factor: 12a 3 6 + 24a 2 6 2 + 12a6 3 

Solution: The greatest common factor (GCF) of 12a 3 6, 24a 2 6 2 and 12a6 3 is 
12ab. Factor out the GCF. 

12a 3 6 + 24a 2 & 2 + 12a6 3 = 12a6 • a 2 - 12ab ■ 2ab + 12ab ■ b 2 



You Try It! 



Factor: 

15s 2 t 4 + 6s 3 t 2 + 9s 2 t 2 



Ylabia 2 + 2ab + b 2 



392 



CHAPTER 6. FACTORING 



Answer: 3s 2 t 2 (5t 2 + 2s + 3) 



Check: Multiply. Distribute the monomial Ylab. 



12ab(a 2 + 2ab + b 2 ) 



12ab ■ a 2 - 12ab ■ 2ab + \2ab ■ b 2 



,2,2 



= YlaTb + 24a z b A + \2ab a 
That's the original polynomial. We have factored correctly. 



□ 



You Try It! 



Factor: 

18p 5 q 4 - 30pV + 42pV 



Answer: 

6p 3 q i (3p 2 -5pq + 7q 2 ) 



Speeding Things Up a Bit 

Eventually, after showing your work on a number of examples such as those in 
Examples 7, 8, and 9, you'll need to learn how to perform the process mentally. 



EXAMPLE 10. Factor each of the following polynomials: (a) 24a; + 32, 
(b) 5a; 3 - 10a; 2 - 10a;, and (c) 2a; 4 y + 2a; 3 y 2 - 6x 2 y 3 . 

Solution: In each case, factor out the greatest common factor (GCF): 

a) The GCF of 24a; and 32 is 8. Thus, 24a; + 32 = 8(3a: + 4) 

b) The GCF of 5a; 3 , 10x 2 , and 10a; is 5a;. Thus: 

5a; 3 - 10a; 2 - 10a; = 5x(a; 2 - 2x - 2) 



c) The GCF of 2a; 4 y, 2a; 3 y 2 , and 6x 2 y 3 is 2a; 2 y. Thus: 



2x i y + 2x 3 y 2 



Qx 2 y 3 



2x y(x + xy — 3y ) 



As you speed things up by mentally factoring out the GCF, it is even more 
important that you check your results. The check can also be done mentally. 
For example, in checking the third result, mentally distribute 2x 2 y times each 
term of x 2 + xy — 3y 2 . Multiplying 2x 2 y times the first term a; 2 produces 2x y, 
the first term in the original polynomial. 

1- 



„2„,3 



2a;^ (x z +xy- 3y z ) = 2x*y + 2x 3 y z - Qx z y 
I t 



Continue in this manner, mentally checking the product of 2a; 2 y with each term 



of x 2 



xy 



3y 2 , making sure that each result agrees with the corresponding 



term of the original polynomial. 

□ 

Remember that the distributive property allows us to pull the GCF out in 
front of the expression or to pull it out in back. In symbols: 



ab + ac = a(b + c) 



ba + ca = (b + c)a 



6. 1 . THE GREATEST COMMON FACTOR 



393 



EXAMPLE 11. Factor: 2x{3x + 2) + 5(3x + 2) 

Solution: In this case, the greatest common factor (GCF) is 3x + 2. 

2x(3x + 2) + 5(3x + 2) = 2x ■ (3x + 2) + 5 • (3x + 2) 
= (2x + 5)(3x + 2) 

Because of the commutative property of multiplication, it is equally valid to 
pull the GCF out in front. 

2x(3x + 2) + 5(3x + 2) = (3x + 2) • 2x + (3x + 2) • 5 
= (3x + 2)(2x + 5) 

Note that the order of factors differs from the first solution, but because of 
the commutative property of multiplication, the order does not matter. The 
answers are the same. 



You Try It! 



Factor: 

3x 2 (4x- 7) + 8(4x - 7) 



Answer: (3a; 2 + 8) (4a; - 7) 



□ 



EXAMPLE 12. Factor: 15a(o + b) - 12 (a + b) 

Solution: In this case, the greatest common factor (GCF) is 3(a + b). 



15a(a + b) - 12(o + i 



3(a + 6)-5a-3(a + 6) -4 
3(a + 6)(5a-4) 



You Try It! 



Factor: 

2Am(m - 2n) + 20(m - 2n) 



Alternate solution: It is possible that you might fail to notice that 15 and 
12 are divisible by 3, factoring out only the common factor a + b. 



15a(a + b) - 12(a + i 



15a- (a + 6)- 12- (a + b) 
(15a- 12) (a + 6) 



However, you now need to notice that you can continue, factoring out 3 from 
both 15a and 12. 



= 3(5a-4)(a + 6) 

Note that the order of factors differs from the first solution, but because of 
the commutative property of multiplication, the order does not matter. The 
answers are the same. 



Answer: 4(6m + 5)(m — 2n) 



□ 



394 



CHAPTER 6. FACTORING 



You Try It! 



Factor by grouping: 
x 2 - 6x + 2x - 12 



Answer: (x + 2)(x — 6) 



Factoring by Grouping 

The final factoring skill in this section involves four-term expressions. The 
technique for factoring a four-term expression is called factoring by grouping. 



EXAMPLE 13. Factor by grouping: x 2 + 8x + 3x + 24 

Solution: We "group" the first and second terms, noting that we can factor 
an x out of both of these terms. Then we "group" the third and fourth terms, 
noting that we can factor 3 out of both of these terms. 

x 2 + 8x + 3x + 24 = x (x + 8) + 3 (x + 8) 



L 



J L 



J 



Now we can factor x + 8 out of both of these terms. 



(x + 3)(x + . 



D 



You Try It! 



Factor by grouping: 
x 2 - 5x - Ax + 20 



Answer: (a; — 4)(.t — 5) 



Let's try a grouping that contains some negative signs. 



EXAMPLE 14. Factor by grouping: x 2 + Ax - 7x - 28 

Solution: We "group" the first and second terms, noting that we can factor x 
out of both of these terms. Then we "group" the third and fourth terms, then 
try to factor a 7 out of both these terms. 

x 2 + Ax - 7x - 28 = x (x + A) + 7 (-x - A) 

{ I t i 

This does not lead to a common factor. Let's try again, this time factoring a 
— 7 out of the third and fourth terms. 

x 2 + Ax - 7x - 28 = x (x + 4) - 7 (x + A) 
t I { i 

That worked! We now factor out a common factor x + A. 

= (x-7){x + A) 



D 



6. 1 . THE GREATEST COMMON FACTOR 



395 



Let's increase the size of the numbers a bit. 



EXAMPLE 15. Factor by grouping: 6a; 2 - 8a; + 9x - 12 

Solution: Note that we can factor 2x out of the first two terms and 3 out of 
the second two terms. 

6a; 2 - 8a; + 9a; - 12 = 2a; (3a; - 4) + 3 (3a; - 4) 
t i t f 

Now we have a common factor 3a; — 4 which we can factor out. 



You Try It! 



Factor by grouping: 
15a; 2 + 9a; + 10a; + 6 



(2a; + 3)(3a;-4) 



Answer: (3a; + 2) (5a; + 3) 



As the numbers get larger and larger, you need to factor out the GCF from 
each grouping. If not, you won't get a common factor to finish the factoring. 



□ 



EXAMPLE 16. Factor by grouping: 24a; 2 - 32a; - 45a; + 60 

Solution: Suppose that we factor 8a; out of the first two terms and —5 out of 
the second two terms. 



You Try It! 



Factor by grouping: 
36a; 2 - 84a; + 15a; - 35 



24a; 2 
t_ 



32a; - 45a; + 60 = 8a; (3a; - 4) - 5 (9a; - 12) 
t t I 



That did not work, as we don't have a common factor to complete the 
factoring process. However, note that we can still factor out a 3 from 9a; — 12. 
As we've already factored out a 5, and now we see can factor out an additional 
3, this means that we should have factored out 3 times 5, or 15, to begin with. 
Let's start again, only this time we'll factor 15 out of the second two terms. 



24a; 2 

t_ 



32a; - 45a; 
_f L_ 



60 



8x (3a; - 4) - 15 (3a; - 4) 



Beautiful! We can now factor out 3a; — 4. 



3a;- 15)(3a;-4) 



Answer: (12a; + 5) (3a; - 7) 



□ 



396 CHAPTER 6. FACTORING 



f> t* t* Exercises ■** •** ■** 

In Exercises 1-6, list all positive divisors of the given number, in order, from smallest to largest. 

1. 42 4. 85 

2. 60 5. 51 

3. 44 6. 63 

In Exercises 7-12, list all common positive divisors of the given numbers, in order, from smallest to 
largest. 

7. 36 and 42 10. 96 and 78 

8. 54 and 30 11. 8 and 76 

9. 78 and 54 12. 99 and 27 



In Exercises 13-18, state the greatest common divisor of the given numbers. 

13. 76 and 8 16. 64 and 76 

14. 84 and 60 17. 24 and 28 

15. 32 and 36 18. 63 and 27 



In Exercises 19-24, use prime factorization to help calculate the greatest common divisor of the given 
numbers. 

19. 600 and 1080 22. 540 and 150 

20. 150 and 120 23. 600 and 450 

21. 1800 and 2250 24. 4500 and 1800 



In Exercises 25-36, find the greatest common factor of the given expressions. 

25. 166 4 and 566 9 28. 24w 3 and 30w 8 

26. 28,s 2 and 36s 4 29. 56ccV and 16a; V 

27. 35^ 2 and 49z 7 30. 356 5 c 3 and 636 4 c 4 



6.1. THE GREATEST COMMON FACTOR 397 

31. 24s 4 i 5 and 16s 3 i 6 34. 8r 7 , 24r 6 , and 12r 5 

32. 10v 4 w 3 and 8v 3 w 4 35. 9a 6 , 6a 5 , and 15a 4 

33. I8y 7 , 45y 6 , and 27y 5 36. 15a 5 , 24a 4 , and 24a 3 

In Exercises 37-52, factor out the GCF from each of the given expressions. 

37. 25a 2 + 10a +20 45. 35s 7 + 49s 6 + 63s 5 

38. 40c 2 + 15c + 40 46. 35s 7 + 56s 6 + 56s 5 

39. 35s 2 + 25s + 45 47. 146 7 + 356 6 + 566 5 

40. 456 2 + 206 + 35 48. 45a; 5 + 81a; 4 + 45a; 3 

41. 16c 3 + 32c 2 + 36c 49. 54y 5 z 3 + 30yV + 36y 3 z 5 

42. 126 3 + 126 2 + 186 50. 42a; 4 ?/ 2 + 42x 3 y 3 + 54x 2 y 4 

43. 42s 3 + 24s 2 + 18s 51. 45s 4 * 3 + 40s 3 t 4 + 15s 2 i 5 

44. 36y 3 + 81y 2 + 36y 52. 20v 6 w 3 + 36v 5 w 4 + 28v 4 w 5 

In Exercises 53-60, factor out the GCF from each of the given expressions. 

53. 7w{2w - 3) - 8(2w - 3) 57. 48a(2a + 5) - 42(2a + 5) 

54. 5s(8s - 1) + 4(8s - 1) 58. 40«(7w - 4) + 72(7w - 4) 

55. 9r(5r - 1) + 8(5r - 1) 59. 56a(2a - 1) - 21(2a - 1) 

56. 5c(4c - 7) + 2(4c - 7) 60. 48r(5r + 3) - 40(5r + 3) 

In Exercises 61-68, factor by grouping. Do not simplify the expression before factoring. 

61. x 2 + 2x - 9x- 18 65. x 2 - 6x - 3x + 18 

62. x 2 + 6a-- 9a;- 54 66. a; 2 - 3a; - 9a; + 27 

63. a; 2 + 3a; + 6a; + 18 67. a; 2 - 9a; + 3a; - 27 

64. a; 2 + 8a; + 7a; + 56 68. a; 2 - 2a; + 7a; - 14 



398 CHAPTER 6. FACTORING 

In Exercises 69-76, factor by grouping. Do not simplify the expression before factoring. 

69. 8a; 2 + 3a; - 56a; - 21 73. 6a; 2 - 7s - 48a; + 56 

70. 4a; 2 + 9a; - 32a; - 72 74. 8a; 2 - 7a; - 72a; + 63 

71. 9a; 2 + 36a; - 5a; - 20 75. 2a; 2 + 12s + 7a; + 42 

72. 7a; 2 + 14s - 8s - 16 76. 7a; 2 + 28a; + 9s + 36 

£»• s*> s*> Answers >** ~*$ •** 

1. {1, 2, 3, 6, 7, 14, 21, 42} 39. 5(7s 2 + 5s + 9) 

3. {1, 2, 4, 11, 22, 44} 41. 4c(4c 2 + 8c + 9) 

5. {1,3,17,51} 43. 6s(7s 2 + 4s + 3) 

7.(1,2,3,6} 45. 7s 5 (5s 2 + 7s + 9) 

9.(1,2,3,6} 47. 76 5 (26 2 + 56 + 8) 

11. {1,2,4} 49. 6y 3 z 3 {9y 2 + 5yz + 6z 2 ) 

13.4 51.5s 2 t 3 (9s 2 + 8st + 3t 2 ) 

15.4 53. (7w-8)(2w-3) 

17. 4 55. (9r + 8)(5r- 1) 

19.120 57. 6(2a + 5)(8a- 7) 

21.450 59. 7(2a- l)(8a-3) 

23. 150 61. (s-9)(s + 2) 

25. 86 4 63. (s + 6)(a; + 3) 

27. 7z 2 65. (s-3)(s-6) 

29. 8x 2 y 4 67. (s + 3)(s-9) 

31. 8s 3 t 5 69. (s- 7)(8a; + 3) 

33. 9y 5 71. (9s-5)(s + 4) 

35. 3a 4 73. (s-8)(6s- 7) 

37. 5(5a 2 + 2a + 4) 75. (2s + 7)(s + 6) 



6.2. SOLVING NONLINEAR EQUATIONS 399 



6.2 Solving Nonlinear Equations 

We begin by introducing a property that will be used extensively in this and 
future sections. 



The zero product property. If the product of two or more numbers equals 
zero, then at least one of the numbers must equal zero. That is, if 

ab = 0, 

then 

a = or 6 = 0. 

Let's use the zero product property to solve a few equations. 



You Try It! 



EXAMPLE 1. Solve for a;: (x + 3) (a; - 5) = Solve for a;: 

Solution: The product of two factors equals zero. 

(x + 3)(x-5) =0 

Hence, at least one of the factors must equal zero. Using the zero product 
property, set each factor equal to zero, then solve the resulting equations for x. 

£ + 3 = or x — 5 = 

x = —3 x = 5 

Hence, the solutions are x = — 3 and x = 5 

Check: Check that each solution satisfies the original equation. 
Substitute —3 for x: Substitute 5 for x: 

(x + 3)(x-5) = (x + 3)(x-5)=0 

(-3 + 3)(-3-5) = (5 + 3)(5-5) = 

(0)(-8) = (8)(0) = 

0=0 0=0 

Because each check produces a true statement, both x = — 3 and x = 5 are 

solutions of (x + 3)(x — 5) = 0. Answer: 7, 2 



The zero product property also works equally well if more than two factors 
are present. For example, if abc = 0, then either a = or b = or c = 0. Let's 
use this idea in the next example. 



□ 



400 



CHAPTER 6. FACTORING 



You Try It! 



Solve for x: 

Qx(x + 4) (5a; + 1) = 



Answer: 0, 



-1/5 



EXAMPLE 2. Solve for x: x(2x + 9) (3a; - 5) = 
Solution: The product of three factors equals zero. 

a;(2a; + 9)(3a;-5) = 

Using the zero product property, set each factor equal to zero, then solve the 
resulting equations for x. 

x = or 



2x + 9 = 


or 


3a; - 5 = 


2a; = -9 




3a; = 5 


9 

x = — 
2 




5 

x = — 

3 



Hence, the solutions are x = 0, x = —9/2, and x = 5/3. We encourage the 
reader to check the solution. 

□ 



Linear versus Nonlinear 

All of the equations solved in previous chapters were examples of what are 
called linear equations. If the highest power of the variable we are solving for 
is one, then the graphs involved are lines. Hence the term, linear equation. 
However, if the power on the variable we are solving for exceeds one, then 
the graphs involved are curves. Hence the term, nonlinear equation. In this 
chapter we will learn how to solve nonlinear equations involving polynomials. 
However, let's first make sure we can recognize the difference between a linear 
and a nonlinear equation. 

Linear versus Nonlinear. Use the following conditions to determine if an 
equation is linear or nonlinear. 

1. If the highest power of the variable we are solving for is one, then the 
equation is linear. 

2. If the highest power of the variable we are solving for is larger than one, 
then the equation is nonlinear. 



You Try It! 



Classify the following 
equation as linear or 
nonlinear: 2x = x 3 — 



EXAMPLE 3. If the instruction is "solve for x, 
equations as linear or nonlinear. 



classify each of the following 



6.2. SOLVING NONLINEAR EQUATIONS 



401 



a) 3x 



7x 



b) x 2 



8x 



Solution: Because the instruction is "solve for a;," to determine whether the 
equation is linear or nonlinear, we identify the largest power of x present in 
the equation. 



a) The highest power of x present in the equation 3x — 5 
Hence, this equation is linear. 



7x is one. 



b) The equation x 2 = 8x contains a power of x higher than one (it contains 
an x 2 ). Hence, this equation is nonlinear. 



Answer: nonlinear 



Now that we can classify equations as either linear or nonlinear, let's in- 
troduce strategies for solving each type, the first of which should already be 
familiar. 

Strategy for solving a linear equation. If an equation is linear, start the 
solution process by moving all terms containing the variable you are solving 
for to one side of the equation, then move all terms that do not contain the 
variable you are solving for to the other side of the equation. 



□ 



You Try It! 



EXAMPLE 4. Solve for 



3a; 



4 -7a; 



Solution: Because the instruction is "solve for x" and we note that the largest 
power of x present is one, the equation 3a: — 5 = 4 — 7x is linear. Hence, the 
strategy is to move all terms containing x to one side of the equation, then 
move all the remaining terms to the other side of the equation. 



Solve for x: 
2x + 4 = 6a; 



3a; - 5 = 4 - 7x 
3a; - 5 + 7x = 4 
3a; + 7x = 4 + 5 



Original equation. 
Add 7x to both sides. 
Add 5 to both sides. 



Note how we have succeeded in moving all terms containing x to one side of the 
equation and all terms that do not contain x to the other side of the equation. 



10a; = 9 Simplify both sides. 

9 
x = — Divide both sides by 10. 

10 

Hence, the solution of 3a; — 5 = 4 — 7x is x = 9/10. Readers are encouraged to 
check this solution. 



Answer: 1/4 



□ 



402 



CHAPTER 6. FACTORING 



You Try It! 



Solve for x: x 2 



-5x 



The situation is much different when the equation is nonlinear. 

Strategy for solving a nonlinear equation. If an equation is nonlinear, 
first move everything to one side of the equation, making one side of the equa- 
tion equal to zero. Continue the solution process by factoring and applying the 
zero product property. 



EXAMPLE 5. Solve for x: 



8x 



Solution: Because the instruction is "solve for a;," and the highest power of 
x is larger than one, the equation x 2 = 8x is nonlinear. Hence, the strategy 
requires that we move all terms to one side of the equation, making one side 
zero. 



x z = 8x 
8a; = 



Original equation. 

Subtract 8a; from both sides. 



Note how we have succeeded in moving all terms to one side of the equation, 
making one side equal to zero. To finish the solution, we factor out the GCF 
on the left-hand side. 



x(a;-8) = 



Factor out the GCF. 



Note that we now have a product of two factors that equals zero. By the zero 
product property, either the first factor is zero or the second factor is zero. 











Hence, the solutions are x = and x = 8. 

Check: Check that each solution satisfies the original equation. 
Substitute for x: Subtitute 8 for x: 



Answer: 0, 



x 2 =8x 

(0) 2 = 8(0) 

= 



x 

(8) 2 
64 



64 



Note that both results are true statements, guaranteeing that both x = and 
x = 8 are solutions of x 2 = 8x. 

□ 



6.2. SOLVING NONLINEAR EQUATIONS 403 



Warning! The following is incorrect! Consider what would happen if we 
divided both sides of the equation x 2 = 8x in Example 5 by a;: 



x z = 8x 
x 2 _ 8x 

X X 






Note that we have lost the second answer found in Example 5, x = 0. This 
example demonstrates that you should never divide by the variable you 
are solving for! If you do, and cancellation occurs, you will lose answers. 

Let's try solving a nonlinear equation that requires factoring by grouping. 



You Try It! 



EXAMPLE 6. Solve for x: 6x 2 + 9x - 8x - 12 = Solve for x: 

5a; 2 — 20a; — 4a; + 16 = 
Solution: Because we are solving for x and there is a power of x larger than 

one, this equation is nonlinear. Hence, the first step is to move everything to 

one side of the equation, making one side equal to zero. Well that's already 

done, so let's factor the left-hand side by grouping. Note that we can factor 

3a; out of the first two terms and —4 out of the second two terms. 

J \ | \ 

6x 2 + 9x - 8x - 12 = 

3a; (2a; + 3) - 4 (2a; + 3) = 
Factor out the common factor 2a; + 3. 

(3a;-4)(2a; + 3) = 

We now have a product of two factors that equals zero. Use the zero product 
property to write: 



3a; - 4 = 


or 


2a; + 3 = 


3a; = 4 




2a; = -3 


4 
X= 3 




3 

X = ~2 



Hence, the solutions are x = 4/3 and x = —3/2. 

Check. Let's use the graphing calculator to check the solution x = 4/3. First, 
store the solution 4/3 in the variable X using the following keystrokes (see the 
first image in Figure 6.1. 



404 



CHAPTER 6. FACTORING 



\T\ 3 EE 



X,T,9,n 




ENTER 



Next, enter the left-hand side of the equation 6a; 2 + 9a; — 8a;— 12 = using the 
following keystrokes. Note that the result in the second image in Figure 6.1 
indicates that the expression 6a; 2 + 9a; — 8a; — 12 equals zero when x = 4/3. 







X,T,0,n 







X,T,0,« 



m 
mm 



ENTER 



Answer: 4/5, 4 



4/3+X 



1 . 333333333 



6*X A 2+9*X-8*X-12 





Figure 6.1: Checking the solution x = 4/3. 

Therefore, the solution x = 4/3 checks. Readers are encouraged to use their 
graphing calculators to check the second solution, x = —3/2. 

□ 



Using the Graphing Calculator 

In this section we will employ two different calculator routines to find the 
solution of a nonlinear equation. Before picking up the calculator, let's first 
use an algebraic method to solve the equation a; 2 = —5a;. The equation is 
nonlinear, so the first step is to move everything to one side of the equation, 
making one side equal to zero. 



x = - 

x 2 + 5a; = 

a;(a; + 5) = 



5 a; 



Nonlinear. Make one side zero. 
Add 5a; to both sides. 
Factor out the GCF. 



Use the zero product property, setting each factor equal to zero, then solving 
the resulting equations for x. 



or x + 5 = 
x = —5 



Hence, the solutions are x = and x 



6.2. SOLVING NONLINEAR EQUATIONS 



405 



We'll now use the calculator to find the solutions of 



-5x. The first 



technique employs the 5:intersect routine on the calculator's CALC menu. 



EXAMPLE 7. Use the 5:intersect utility on the graphing calculator to 
solve the equation x 2 = — 5x for x. 

Solution: Load the left-hand side of x 2 = —5x in Yl and the right-hand 
side in Y2 (see Figure 6.2). Selecting 6:ZStandard from the ZOOM menu 
produces the graphs shown in the image on the right in Figure 6.2. 



You Try It! 



Use the 5: intersect utility 
on the graphing calculator to 
solve the equation x 2 = Ax 
for x. 



Plotl PlotE Plots 






E / 


WiBX A 2 






; / 


^.VeB-5+X 






:/ 








•J. 






We = 








\Vs= 








\V?= 









Figure 6.2: Sketch the graphs of each side of the equation x 2 



-hx. 



Note that the graph of y = x 2 is a parabola that opens upward, with vertex 
(turning point) at the origin. This graph reveals why the equation x 2 = —5x 
is called a nonlinear equation (not all the graphs involved are lines). Next, the 
graph of y = —5x is a line with slope —5 and y-intercept at the origin. 

The two graphs obviously intersect at the origin, but it also appears that 
there may be another point of intersection that is off the screen. Let's increase 
Ymax in an attempt to reveal the second point of intersection. After some 
experimentation, the settings shown in the first image in Figure 6.3 reveal both 
points of intersection. Pushing the GRAPH button produces the image on the 
right in Figure 6.3. 



WINDOW 
Xnin=-10 
Xnax=10 
Xscl=l 
Ymin=-20 
Vnax=5@ 
Yscl=10 




Figure 6.3: Adjust the WINDOW parameters to reveal both points of inter- 
section. 



To find the solutions of the equation x 2 
of the points where the graphs of y = 



-5x, we must find the coordinates 



and y 



-5x intersect. The x- 



coordinate of each point of intersection will be a solution of the equation x 2 
—5x. 



406 



CHAPTER 6. FACTORING 



• Start by selecting 5:intersect from the CALC menu. When prompted 
for the "First curve?" , press ENTER. When prompted for the "Second 
curve?", press ENTER. When prompted for a "Guess," press ENTER. 
The result is the point (0, 0) shown in the image on the left in Figure 6.4. 

• Repeat the process a second time. Select 5:intersect from the CALC 
menu. When prompted for the "First curve?" , press ENTER. When 
prompted for the "Second curve?" , press ENTER. When prompted for a 
"Guess," use the left-arrow key to move the cursor closer to the leftmost 
point of intersection, then press ENTER. The result is the point (—5, 25) 
shown in the image on the right in Figure 6.4. 



\ 




Intersection^ 
K=0 / 


Y=0 





■V 


Intersection' 


7=16 





■J 


Intersection 
K=0 


V=0\ 





■J 


Intersection 
K=-£ 


Y=££ 



Figure 6.4: Use the 5:intersect utility to find the points of intersection. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.5). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.5). 

• Label each graph with its equation (see Figure 6.5). 

• Drop dashed vertical lines through each point of intersection. Shade and 
label the x- values of the points where the dashed vertical line crosses the 
x-axis. These are the solutions of the equation x 2 = — 5x (see Figure 6.5). 

Hence, the solutions of x 2 = — 5x are x = —5 and x = 0. Note now these 
match the solutions found using the algebraic technique. 

□ 



Before demonstrating a second graphing calculator technique for solving 
nonlinear equations, let's take a moment to recall the definition of a zero of a 
function, which was first presented in Chapter 5, Section 3. 



Zeros and ^-intercepts. The points where the graph of / crosses the x-axis 
are called the x-intercepts of the graph of /. The x-value of each x-intercept 
is called a zero of the function /. 



6.2. SOLVING NONLINEAR EQUATIONS 



407 



y = — 5x 




Figure 6.5: Reporting your graphical solution on your homework. 



We'll now employ the 2:zero utility from the CALC menu to find the solutions 
of the equation x 2 = — 5x. 



You Try It! 



EXAMPLE 8 

the equation x 2 

Solution: First, make one side of the equation equal to zero 



Use the 2:zero utility on the graphing calculator to solve 
— 5x for x. 



Use the 2:zero utility on the 
graphing calculator to solve 
the equation x 2 = Ax for x. 



x 
5x 



-5x 







Make one side zero. 
Add 5a; to both sides. 



To determine the values of x that make x 2 + 5x = 0, we must locate the points 
where the graph of f(x) = x 2 + 5x crosses the a>axis. These points are the 
ai-intercepts of the graph of / and the a;- values of these points are the zeros of 
the function /. 

Load the function f(x) = x 2 + 5x in Yl, then select 6:ZStandard to 
produce the image in Figure 6.6. Note that the graph of / has two ai-intercepts, 
and the a;- values of each of these points are the zeros of the function /. 



Plotl PlotE Plots 
WiBX A 2+5*X 
-■■Vi = 

Wj = 

We = 






J 







It's often easier to find the so- 
lutions of a nonlinear equa- 
tion by making one side zero 
and identifying where the 
graph of the resulting func- 
tion crosses the a>axis. 



Figure 6.6: Sketch the graph of p(x) 



hx. 



Select 2:zero from the CALC menu (see Figure 6.7). 



408 



CHAPTER 6. FACTORING 



• The calculator responds by asking for a "Left Bound?" Use the left-arrow 
key to move the cursor so that it lies to the left of the ^-intercept near 
(—5, 0) (see the second image in Figure 6.7), then press the ENTER key. 

• The calculator responds by asking for a "Right Bound?" Move the cursor 
so that is slightly to the right of the x-intercept near (—5, 0) (see the third 
image Figure 6.7), then press the ENTER key. 



lilla 



jBualue 
Z: zero 
3: rdniFiUFi 
4: maximun 
5: inter-sect 
6:dy^dx 
7:^fCx>dx 





1 


Left Bound? 
K= "£.319149 


Y=l.fi97fi007 



Y1=!T£+E*H 




RiSht Bound? 

K—H.SHOHSl 



V="1.H93BH9 



Figure 6.7: Setting the left and right bounds when using the 2:zero utility to 
find the ^-intercepts of the graph of f{x) = x 2 + 5x. 



• The calculator responds by asking for a "Guess?" Note the two triangular 
marks near the top of the viewing window in the first image in Figure 6.8 
that mark the left- and right-bounds. As long as you place the cursor 
so that the rvalue of the cursor location lies between these two marks, 
you've made a valid guess. Because the cursor already lies between these 
two marks, we usually leave it where it is and press the ENTER key. 



Y1=!T£+E*H 
4 




Gut::? 

K=-4.fiH0B£i V="1.H93BH9 





t 


K=-£ 


V=0 



Figure 6.8: Setting the right bound and making a guess. 



After making your guess and pressing the ENTER key, the calculator pro- 
ceeds to find an approximation of the ^-intercept that lies between the left- and 
right-bounds previously marked (see the second image in Figure 6.8). Hence, 
this ^-intercept is (—5, 0), making —5 a zero of f{x) = x 2 + 5x and a solution 
of the equation x 2 + 5x = 0. 

We'll leave it to our readers to repeat the 2:zero process to find the second 
zero at the origin. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 



6.2. SOLVING NONLINEAR EQUATIONS 



409 



• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.9). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.9). 

• Label each graph with its equation (see Figure 6.9). 

• Drop dashed vertical lines through each ^-intercept. Shade and label 
the a:- values of each x-intercept. These are the solutions of the equation 



-5x (see Figure 6.9). 




Answer: 







Z*K* 
K=0 


V=0 



Figure 6.9: Reporting your graphical solution on your homework. 

Hence, the solutions of x 2 = —5a; are x = —5 and x = 0. Note how 
nicely this agrees with the solutions found using the algebraic technique and 
the solutions found using the 5:intersect utility in Example 7. 



...\l 




Z*K0 


V=0 



□ 



410 CHAPTER 6. FACTORING 



*».*».**. Exercises ■** •** * 

In Exercises 1-8, solve the given equation for x. 

1. (9a; + 2) (8a; + 3) =0 5. -9a;(9a; + 4) = 

2. (2x-5)(7x-4) = 6. 4a;(3a; - 6) = 

3. x(4x + 7)(9» - 8) = 7. (a; + 1)(» + 6) = 

4. ar(9a;-8)(3a; + l) = 8. (a? - 4)(x - 1) = 

In Exercises 9-18, given that you are solving for x, state whether the given equation is linear or 
nonlinear. Do not solve the equation. 

9. x 2 + 7x = 9x + 63 14. 4x 2 = -7x 

10. x 2 + 9x = ix + 36 15. 3x 2 + 8x = -9 

11. 6a; - 2 = 5a; - 8 16. 5a; 2 - 2a; = -9 

12. -5a; + 5 = -6a; - 7 17. -3a; + 6 = -9 

13. 7a; 2 = -2a; 18. 8a; - 5 = 3 



In Exercises 19-34, solve each of the given equations for x. 

19. 3a; + 8 = 9 27. 9a; + 2 = 7 

20. 3a; + 4 = 2 28. 3a; + 2 = 6 

21. 9a; 2 = -x 29. 9a; 2 = 6a; 

22. 6a; 2 = 7a; 30. 6a; 2 = -14a; 

23. 3a; + 9 = 8x + 7 31. 7a; 2 = -4x 

24. 8a; + 5 = 6a; + 4 32. 7a; 2 = -9a; 

25. 8a; 2 = -2a; 33. 7a; + 2 = 4a; + 7 

26. 8a; 2 = 18x 34. 4x + 3 = 2x + 8 



6.2. SOLVING NONLINEAR EQUATIONS 411 

In Exercises 35-50, factor by grouping to solve each of the given equations for x. 

35. 63a; 2 + 56a; + 54a; + 48 = 43. x 2 + 6x - 11a; - 66 = 

36. 27a; 2 + 36a; + 6a; + 8 = 44. a; 2 + 6a; - 2x - 12 = 

37. 16a; 2 - 18a; + 40a; - 45 = 45. 15a; 2 - 24a; + 35a; - 56 = 

38. 42a; 2 - 35a; + 54a; - 45 = 46. 12a; 2 - 10a; + 54a; - 45 = 

39. 45a; 2 + 18a; + 20a; + 8 = 47. a; 2 + 2a; + 9a; + 18 = 

40. 18a; 2 + 21a; + 30a; + 35 = 48. a; 2 + 8a; + 4ai + 32 = 

41. a; 2 + 10a; + 4a; + 40 = 49. a; 2 + 4a; - 8a; - 32 = 

42. a; 2 + 11a; + 10a; + 110 = 50. a; 2 + 8a; - 5a; - 40 = 



In Exercises 51-54, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the 
results found using graphing calculator as shown in Example 7. 

51. a; 2 = -Ax 53. a; 2 = 5a; 

52. a; 2 = 6a; 54. a; 2 = -6a; 



In Exercises 55-58, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results 
found using graphing calculator as shown in Example 8. 

55. a; 2 + 7a; = 57. a; 2 - 3a; = 

56. a; 2 - 8a; = 58. a; 2 + 2a; = 



£*• £*< £*■ Answers •*$ •** **$ 



2 3 4 

1. x = — , — 5. x = 0, — 

9' 8 '9 



7 8 
3. x = 0, --, - 

' 4' 9 



7. x = -1,-6 
9. Nonlinear 



11. 


Linear 


13. 


Nonlinear 


15. 


Nonlinear 


17. 


Linear 


19. 


1 
3 


21. 


1 

x = Q, — 

9 


23. 


2 

5 


25. 


1 
x = 0,-- 


27. 


5 
9 


29. 


2 


31. 


4 
* = 0, -- 


33. 


5 
3 



412 CHAPTER 6. FACTORING 

68 
35. x = --, -- 

7' 9 

59 
37. x= --, - 

2' 8 

4 2 
39. x = --, -- 

9' 5 



41. x = -4, -10 

43. x= 11, -6 

7 8 
45. x = — , - 

3' 5 
47. x= -9, -2 

49. x = 8, -4 

51. x= -4,0 

53. x = 0,5 

55. x= -7,0 

57. x = 0,3 



6.3. FACTORING AX 2 + BX + C WHEN A = 1 



413 



6.3 Factoring ax 2 + bx + c when a = 1 

In this section we concentrate on learning how to factor trinomials having the 
form ax 2 + bx + c when a = 1. The first task is to make sure that everyone can 
properly identify the coefficients a, b, and c. 



You Try It! 



EXAMPLE 1. Compare x 2 — 8x — 9 with the form ax 2 + bx + c and identify Compare 2a; 2 + 5a; — 3 with 



the coefficients a, 6, and c. 

Solution: Align the trinomial x 2 — 8x — 9 with the standard form ax 2 + bx + c, 
then compare coefficients. Note that the understood coefficient of x 2 is 1. 



the form ax 2 + bx + c and 
identify the coefficients a, b, 
and c. 



ax 



bx+c 



lx z - 8a;-9 



We see that a = 1, b = — 8, and c = — 9. Because the leading coefficient is 1, 
this is the type of trinomial that we will learn how to factor in this section. 



Answer: 
a = 2,b 



5, c 



□ 



EXAMPLE 2. Compare -40 + 6x 2 
identify the coefficients a, b, and c. 



x with the form ax 2 + bx + c and 



Solution: First, arrange —40 + 6a; 2 — x in descending powers of x, then align 



it with the standard form 



-bx + c and compare coefficients. Note that the 



understood coefficient of a; is — 1. 



You Try It! 



Compare 3a; + 9 — 7a; 2 with 
the form ax 2 + bx + c and 
identify the coefficients a, 6, 
and c. 



6a; 2 



bx + c 
la; -40 



We see that a = 6, b = — 1, and c = —40. Because the leading coefficient is 
6, we will have to wait until Factoring aa; 2 + bx + c when a/ 1 on page 427 
before learning how to factor this trinomial. 



Answer: 
a = —7, 



3, c = 9 



□ 



In this section, the leading coefficient must equal 1. Our work in this 
section will focus only on trinomials of the form x 2 + bx + c, that is, the form 



ax 



bx + c where a = 1. 



The ac-Method 

We are now going to introduce a technique called the ac-method (or ac-test) 
for factoring trinomials of the form aa; 2 +bx + c when a = 1. In the upcoming 



414 



CHAPTER 6. FACTORING 



section Factoring ax 2 + bx + c when a^l on page 427, we will see that this 
method can also be employed when o / 1, with one minor exception. But 
for the remainder of this section, we focus strictly on trinomials whose leading 
coefficient is 1. 

Let's begin by finding the following product: 



(x + 12) (a; — 4) = x(x — 4) + 12(cc — 4) Apply the distributive property. 

Distribut 
Simplify. 



X — 4a: + 12 x — 48 Distribute again. 



■ 8a; - 48 



Now, can we reverse the process? That is, can we start with x 2 + 8x — 48 and 
place it in its original factored form (x + 12) (x — 4) ? The answer is yes, if we 
apply the following procedure. 



The ac-method. Compare the given polynomial with the standard form 
ax 2 + bx + c, determine the coefficients a, b, and c, then proceed as follows: 

1. Multiply the coefficients a and c and determine their product ac. List all 
the integer pairs whose product equals ac. 

2. Circle the pair in the list produced in step 1 whose sum equals b, the coef- 



ficient of the middle term of ax 



+ c. 



3. Replace the middle term bx with a sum of like terms using the circled pair 
from step 2. 

4. Factor by grouping. 

5. Check the result using the FOIL shortcut. 



You Try It! 



Factor: x 2 + 11a; + 28 



Let's follow the steps of the ac-method to factor x 2 + 8x — 48. 



EXAMPLE 3. Factor: 



8a; - 48. 



Solution: Compare x 2 + 8a; — 48 with ax 2 + bx +c and identify a = 1, b = 8, and 
c = —48. Note that the leading coefficient is a = 1. Calculate ac. Note that 
ac = (1)(— 48), so ac = —48. List all integer pairs whose product is ac = —48. 



1. 


-48 


-1,48 


2. 


-24 


-2,24 


3. 


-16 


-3,16 


4- 


-12 


-4,12 


6. 


-8 


-6,8 



6.3. FACTORING AX 2 + BX + C WHEN A = 1 



415 



Circle the ordered pair whose sum is b 



-48 
-24 
-16 
-12 



-1,48 
-2,24 
-3,16 



-4,12 



-6,8 



x 2 +8x - 


-48 = x 2 -ix + 12a; -48 


2 + 8x- 


- 48 = x(x - 4) + 12(sc - 4) 




= (a: + 12)(a:-4) 



Replace the middle term 8x with a sum of like terms using the circled pair 
whose sum is 8. 



Factor by grouping. 



Use the FOIL shortcut to mentally check your answer. To determine the prod- 
uct (x + 12) (a; — 4), use these steps: 

• Multiply the terms in the "First" positions: x 2 . 

• Multiply the terms in the "Outer" and "Inner" positions and combine 
the results mentally: —Ax + Ylx = 8x. 

• Multiply the terms in the "Last" positions: —48. 
That is: 



(a; + 12)(a:-4) 



F 

,,2 



o 

Ax 



I 
Ylx 



L 

48 



Combining like terms, (x + 12) (x — A) 



8a; — 48, which is the original 



trinomial, so our solution checks. Note that if you combine the "Outer" and 
"Inner" products mentally, the check goes even faster. 



Answer: (x + A)(x + 7) 



□ 



Some readers might ask "Is it a coincidence that the circled pair —A, 12 



seemed to 'drop in place' in the resulting factoriztion (x + 12)(x — 4)?" Before 
we answer that question, let's try another example. 



You Try It! 



EXAMPLE 4. Factor: 



9x - 36. 



Factor: 



10a; - 24 



Solution: Compare a; 2 — 9a; — 36 with ax 2 + bx + c and note that a = 1, b = —9, 
and c = —36. Calculate ac = (1)(— 36), so ac = —36. 

At this point, some readers might ask "What if I start listing the ordered 
pairs and I see the pair I need? Do I need to continue listing the remaining 
pairs? 



416 



CHAPTER 6. FACTORING 



Answer: (a; + 12) (x - 2) 



The answer is "No." In this case, we start listing the integer pairs whose 
product is ac = —36, but are mindful that we need an integer pair whose sum 
is b = — 9. The integer pair 3 and —12 has a product equaling ac = —36 and a 
sum equaling b = — 9. 

1,-36 

2,-18 



-12 



Note how we ceased listing ordered pairs the moment we found the pair we 
needed. Next, replace the middle term —9a; with a sum of like terms using the 
circled pair. 



Factor by grouping. 



x 2 -9x - 36 = x 2 +3x - Ylx - 36 



x 2 -9x - 36 = x(x + 3) - 12(ar + 3) 



= (x- 12)(x + 3) 
Use the FOIL shortcut to check your answer. 



(x + 3)(x- 12) 



F 

^2 



o 

Ylx 



I 

3.7' 



L 
36 



Combining like terms, (x + 3) (a; — 12) 
Our solution checks. 



9a; — 36, the original trinomial. 



□ 



You Try It! 



Factor: x 2 - Ylx + 35 



-12 



Speeding Things Up a Bit 

Readers might again ask "Is it a coincidence that the circled pair 
seemed to 'drop in place' in the resulting factoriztion (x — 12) (a: + 3) ?" The 
answer is "No," it is not a coincidence. Provided the leading coefficient of the 
trinomial ax 2 + bx + c is a = 1, you can always "drop in place" the circled pair 
in order to arrive at the final factorization, skipping the factoring by grouping. 
Some readers might also be asking "Do I really have to list any of those 
ordered pairs if I already recognize the pair I need?" The answer is "No!" If 
you see the pair you need, drop it in place. 



EXAMPLE 5. Factor: 



5a; - 24. 



Solution: Compare x 2 — 5a; — 24 with ax 2 + bx + c and note that a = 1, b = — 5, 
and c = —24. Calculate ac = (1)(— 24), so ac = —24. Now can you think of an 
integer pair whose product is ac = —24 and whose sum is 6 = —5? For some, 
the required pair just pops into their head: —8 and 3. The product of these 



6.3. FACTORING AX 2 + BX + C WHEN A=l 417 

two integers is —24 and their sum is —5. "Drop" this pair in place and you are 
done. 

x 2 -5x-24 = {x-8)(x + 3) 
Use the FOIL shortcut to check your answer. 

FOIL 
(x-8)(x + 3) = x 2 + 3x - 8x - 24 

Combining like terms, (a: — 8)(x + 3) = x 2 — 5x — 24, the original trinomial. 

Our solution checks. Answer: (x — 7)(x — 5) 



The "Drop in Place" technique of Example 5 allows us to revise the ac- 
method a bit. 



Revised ac-method. Compare the given polynomial with the standard form 
ax 2 + bx + c, determine the coefficients a, b, and c, then determine a pair of 
integers whose product equals ac and whose sum equals b. You then have two 
options: 

1. Write the middle term as a product of like terms using the ordered pair 
whose product is ac and whose sum is b. Complete the factorization process 
by factoring by grouping. 

2. (Only works if a = 1.) Simply "drop in place" the ordered pair whose 
product is ac and whose sum is b to complete the factorization process. 
Note: We'll see in Factoring ax 2 + bx + c when a ^ 1 on page J^21 why this 
"drop in place" choice does not work when a/1. 



Readers are strongly encouraged to check their factorization by determining 
the product using the FOIL method. If this produces the original trinomial, 
the factorization is correct. 



Nonlinear Equations Revisited 

The ability to factor trinomials of the form ax 2 + bx + c, where a = 1, increases 
the number of nonlinear equations we are now able to solve. 



□ 



You Try It! 



EXAMPLE 6. Solve the equation x 2 = 2x + 3 both algebraically and Solve the equation 
graphically, then compare your answers. x 2 = —3a; + 4 both 

algebraically and graphically, 
then compare your answers. 



418 



CHAPTER 6. FACTORING 



Solution: Because there is a power of x larger than one, the equation is 
nonlinear. Make one side zero. 



x 
x 2 - 2x 
-2x- 3 



2x 

3 





Original equation. 

Subtract 2x from both sides. 

Subtract 3 from both sides. 



Compare x 2 — 2x — 3 with ax 2 + bx + c and note that a = 1, b = —2 and c = —3. 
We need an integer pair whose product is ac = —3 and whose sum is 6 = — 2. 
The integer pair 1 and —3 comes to mind. "Drop" these in place to factor. 



0+1)0-3) = 



Factor. 



We have a product that equals zero. Use the zero product property to complete 
the solution. 



or 



x + 1 =0 

x = -l 

Thus, the solutions of x 2 = 2x + 3 are x 



x-3 = 
x = 3 
1 and x = 3. 



Graphical solution. Load each side of the equation x 2 = 2x + 3 into the 
Y= menu of your graphing calculator, y = x 2 in Yl, y = 2x + 3 in Y2 (see 
Figure 6.10). Select 6:ZStandard from the ZOOM menu to produce the image 
at the right in Figure 6.10. 



Plotl PlotE Plots 
^V£B2*X+3 




Figure 6.10: Sketch y = x 2 and y = 2x + 3. 



One of the intersection points is visible on the left, but the second point of 
intersection is very near the top of the screen at the right (see Figure 6.10). 
Let's extend the top of the screen a bit. Press the WINDOW button and make 
adjustments to Ymin and Ymax (see Figure 6.11), then press the GRAPH 
button to adopt the changes. 

Note that both points of intersection are now visible in the viewing window 
(see Figure 6.11). To find the coordinates of the points of intersection, select 
5:intersect from the CALC menu. Press the ENTER key to accept the "First 
curve," press ENTER again to accept the "Second curve," then press ENTER 
again to accept the current position of the cursor as your guess. The result is 
shown in the image on the left in Figure 6.12. Repeat the process to find the 
second point of intersection, only when it comes time to enter your "Guess," use 
the right-arrow key to move the cursor closer to the second point of intersection 
than the first. 



6.3. FACTORING AX 2 + BX + C WHEN A = 1 



419 



l.i.lIHDOI.1 




Xmin= 


-10 


Xmax= 


10 


y.scl = 


1 


Vnin= 


-5 


Vmax= 


15 


Yscl = 


1 


l¥.res= 


I 




Figure 6.11: Adjusting the view. 







Intersection 
K=-l / 


V=l 





\v 


Intersection 
K=3 / 


Y=9 



Figure 6.12: Use 5:intersect from the CALC menu to find points of intersection. 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.13). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.13). 

• Label each graph with its equation (see Figure 6.13). 

• Drop dashed vertical lines through each point of intersection. Shade and 
label the x-values of the points where the dashed vertical line crosses 
the x-axis. These are the solutions of the equation x 2 = 2x + 3 (see 
Figure 6.13). 

Finally, note how the graphical solutions of a: 2 = 2x + 3, namely x = — 1 and 
x = 3, match the solutions found using the algebraic method. This is solid 
evidence that both methods of solution are correct. However, it doesn't hurt 
to check the final answers in the original equation, substituting —1 for x and 
3 for x. 



(-1) 2 



1 



22 + 3 

2(-l) + 3 
-2 + 3 



and 



x 

(3) 2 
9 



2x + 3 

2(3) + 3 
6 + 3 



Because the last two statements are true statements, the solutions x 
x = 3 check in the original equation x 2 = 2x + 3. 



-1 and 



Answer: —4, 1 



TV 




Intersection 
K=-4 


Y=lfiX 





\J 


Intersection 
K=l 


X-IK 



□ 



420 



CHAPTER 6. FACTORING 




Figure 6.13: Reporting your graphical solution on your homework. 



You Try It! 



Solve the equation 
x 2 - 2\x + 90 = both 
algebraically and graphically, 
then compare your answers. 



EXAMPLE 7. Solve the equation x 2 
graphically, then compare your answers. 



4x — 96 = both algebraically and 



Solution: Because there is a power of x larger than one, the equation x 2 — 
Ax — 96 = is nonlinear. We already have one side zero, so we can proceed with 
the factoring. Begin listing integer pairs whose product is ac = —96, mindful 
of the fact that we need a pair whose sum is b = —4. 



1,-96 
2,-48 
3,-32 
4,-24 
6,-16 



-12 



Note that we stopped the listing process as soon as we encountered a pair 
whose sum was b = —4. "Drop" this pair in place to factor the trinomial. 



x z - Ax - 96 
(x + 8)(x- 12) 



Original equation. 
Factor. 



6.3. FACTORING AX 2 + BX + C WHEN A = 1 



421 



We have a product that equals zero. Use the zero product property to complete 
the solution. 



x + : 







a; -12 = 
cc = 12 



Thus, the solutions of x 2 



Ax 



are x 



and x = 12. 



Graphical solution. Load the equation y = x 2 — Ax — 96 in Yl in the Y= 
menu of your graphing calculator (see Figure 6.14). Select 6:ZStandard from 
the ZOOM menu to produce the image at the right in Figure 6.14. 



Plotl PlotE Plots 
WiBX A 2-4*X-96 
-■■Vi = 



Figure 6.14: Sketch the graph of y 



Ax - 96. 



When the degree of a polynomial is two, we're used to seeing some sort of 
parabola. In Figure 6.14, we saw the graph go down and off the screen, but we 
did not see it turn and come back up. Let's adjust the WINDOW parameters 
so that the vertex (turning point) of the parabola and both ^-intercepts are 
visible in the viewing window. After some experimentation, the settings shown 
in Figure 6.15 reveal the vertex and the ^-intercepts. Press the GRAPH button 
to produce the image at the right in Figure 6.15. 



WINDOW 
Xnin=-15 
Xnax=2@ 
Xscl=5 
Ymin=-150 
Vnax=15@ 
Yscl=50 




Figure 6.15: Adjusting the view. 



Note that both ^-intercepts of the parabola are now visible in the viewing 
window (see Figure 6.15). To find the coordinates of the x-intercepts, select 
2:zero from the CALC menu. Use the left- and right-arrow keys to move the 
cursor to the left of the first x- intercept, then press ENTER to mark the "Left 
bound." Next, move the cursor to the right of the first x-intercept, then press 
ENTER to mark the "Right bound." Press ENTER to accept the current 
position of the cursor as your "Guess." The result is shown in the image on 
the left in Figure 6.16. Repeat the process to find the coordinates of the second 
x-intercept. The result is show in the image on the right in Figure 6.16. 



422 



CHAPTER 6. FACTORING 



\ 






V=0 



\l 






V=0 



Figure 6.16: Use 2:zero from the CALC menu to find the ^-intercepts. 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.17). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.17). 

• Label the graph with its equation (see Figure 6.17). 

• Drop dashed vertical lines through each x-intercept. Shade and label the 
x- values of the points where the dashed vertical line crosses the x-axis. 
These are the solutions of the equation x 2 — Ax — 96 = (see Figure 6.17). 




150 



-150 



y = x 



2 - Ax - 96 



-»■ x 



20 




Figure 6.17: Reporting your graphical solution on your homework. 



Finally note how the graphical solutions of x 2 — Ax — 96 = 0, namely x = —8 
and x = 12, match the solutions found using the algebraic method. This is 
solid evidence that both methods of solution are correct. However, it doesn't 



6.3. FACTORING AX 2 + BX + C WHEN A = 1 



423 



hurt to check the final answers in the original equation, substituting —8 for x Answer: 6, 15 
and 12 for x. 



x z - Ax - 96 = 



(-8) 2 -4(- 



64 + 32 



and x 2 - Ax - 96 = 

(12) 2 -4(12) -96 = 

144 - 48 - 96 = 




Because the last two statements are true statements, the solutions x = —8 and 
x = 12 check in the original equation x 2 — Ax — 96 — 0. 




□ 



424 CHAPTER 6. FACTORING 

**> t* t* Exercises ■** ■*$ •** 

In Exercises 1-6, factor the quadratic polynomial. 

1. a; 2 + 7a; -18 4. x 2 + Ylx + 27 

2. x 2 + 18a; + 80 5. a; 2 + 14a; + 45 

3. a; 2 -10a; + 9 6. a; 2 + 9a; + 20 



In Exercises 7-12, factor the quadratic polynomial. 

7. a; 2 - 16a; + 39 10. a; 2 - 22a; + 57 

8. a; 2 - 16a; + 48 11. a; 2 - 25a; + 84 

9. a; 2 - 26a; + 69 12. a; 2 + 13a; - 30 



In Exercises 13-18, factor the quadratic polynomial. 

13. a; 2 - 13a; + 36 16. a; 2 - 17a; + 66 

14. a; 2 + x -12 17. a; 2 - 4a; - 5 

15. a; 2 + 10a; + 21 18. a; 2 - 20a; + 99 



In Exercises 19-24, use an algebraic technique to solve the given equation. 

19. x 2 = -7x + 30 22. a; 2 = x + 72 

20. a; 2 = -2a; + 35 23. x 2 = -15x - 50 

21. a; 2 = -lice- 10 24. a; 2 = -7x - 6 

In Exercises 25-30, use an algebraic technique to solve the given equation. 

28. 80 = a; 2 - 16a; 

29. 56 = a; 2 + 10a; 

30. 66 = a; 2 + 19a; 



25. 


60 = 


a; 2 + 11a; 


26. 


-92 




= ar - 


-27a; 


27. 


-11 




= X - 


- 12a; 



6.3. FACTORING AX 2 + BX + C WHEN A=l 425 

In Exercises 31-36, use an algebraic technique to solve the given equation. 

31. x 2 + 20 = -12a; 34. x 2 + 6 = 5a; 

32. x 2 - 12 = 11a; 35. x 2 + 8 = -6x 

33. x 2 - 36 = 9a; 36. a; 2 + 77 = 18a; 



In Exercises 37-40, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the 
results found using graphing calculator as shown in Example 6. 

37. x 2 = x + 12 39. x 2 + 12 = 8a: 

38. x 2 = 20 - x 40. x 2 + 7 = 8x 



In Exercises 41-44, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results 
found using graphing calculator as shown in Example 7. 

41. x 2 -6x- 16 = 43. x 2 + 10a; - 24 = 

42. x 2 + 7a;- 18 = 44. x 2 - 9x - 36 = 



it- it- it- Answers •*$ •** •** 

15. (a; + 3)(a; + 7) 
17. (x + l)(x-5) 
19. x = 3, -10 
21. x = -1,-10 
23. x = -5,-10 
25. x = 4, -15 
27. x = 1,11 
29. a; = 4, -14 



1. 


(x-2)(x + 9) 


3. 


(x- l)(x-9) 


5. 


(a; + 5)(a; + 9) 


7. 


(x-3)(x- 13) 


9. 


(x-3)(x-23) 


11. 


(x-4)(x-21) 


13. 


(x-4)(x-9) 



426 CHAPTER 6. FACTORING 

31. a; = -2,-10 39. a; = 2, 6 

33 ^ = - 3 ' 12 41., = 8, -2 

35. x = -2,-4 

43. x = -12,2 
37. x = -3,4 



6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 



427 



6.4 Factoring ax 2 + bx + c when o/l 

In this section we continue to factor trinomials of the form ax 2 + bx + c. In 
the last section, all of our examples had a = 1, and we were able to "Drop in 
place" our circled integer pair. However, in this section, a/ 1, and we'll soon 
see that we will not be able to use the "Drop in place" technique. However, 
readers will be pleased to learn that the ac-method will still apply. 



EXAMPLE 1. Factor: 2a; 2 - 7x - 
Solution: We proceed as follows: 



15. 



1. Compare 2a; 2 — 7a; — 15 with aa; 2 + bx + c and identify a = 2, b = —7, 
and c = —15. Note that the leading coefficient is a = 2, so this case is 
different from all of the cases discussed in Section 6.3. 

2. Calculate ac. Note that ac = (2)(— 15), so ac = —30. 

3. List all integer pairs whose product is ac = —30. 



1. 


-30 


-1,30 


2, 


-15 


-2,15 


3, 


-10 


-3,10 


■5. 


-6 


-5,6 



You Try It! 



Factor: 3a; 2 + 13a; + 14 



4. Circle the ordered pair whose sum is 



-7. 



1. 

2, 


-30 

-15 


-1,30 
-2,15 


:'.. 


-10 


-3,10 


5, 


-6 


-5,6 



Note that if we "drop in place" our circled ordered pair, (x + 3)(x— 10) 7^ 
2a; 2 — 7x — 15. Right off the bat, the product of the terms in the "First" 
position does not equal 2a; 2 . Instead, we break up the middle term of 
2a; 2 — 7a; — 15 into a sum of like terms using our circled pair of integers 
3 and -10. 

2x 2 -7a; - 15 = 2a; 2 +3a; - 10a; - 15 

Now we factor by grouping. Factor x out of the first two terms and —5 
out of the second two terms. 



x (2a; + 3) - 5 (2a; + 3) 



Now we can factor out (2a; + 3). 



(a;-5)(2a; + 3) 



428 



CHAPTER 6. FACTORING 



Answer: (x + 2)(3x + 7) 



6. Use the FOIL shortcut to mentally check your answer. To multiply 
(x — 5) (2a; + 3), use these steps: 

• Multiply the terms in the "First" positions: 2x 2 . 

• Multiply the terms in the "Outer" and "Inner" positions and com- 
bine the results mentally: 3x — Wx = —7x. 

• Multiply the terms in the "Last" positions: —15. 



That is: 



(x-5)(2sc + 3) 



F 

2x 2 



O 

3.r 



10a; 



L 

15 



Combining like terms, (x — 5)(2x + 3) = 2a; 2 — 7x — 15, which is the 
original trinomial, so our solution checks. Note that if you combine the 
"Outer" and "Inner" products mentally, the check goes even faster. 



□ 



You Try It! 



Factor: 2a: 2 - 9a; + 10 



Speeding Things Up a Bit 

Some readers might already be asking "Do I really have to list all of those 
ordered pairs if I already see the pair I need?" The answer is "No!" If you see 
the pair you need, use it to break up the middle term of the trinomial as a sum 
of like terms. 



EXAMPLE 2. Factor: 3a; 2 - 7x - 6. 

Solution: Compare 3a; 2 — 7a; — 6 with ax 2 + bx + c and note that a = 3, b = —7, 
and c = —6. Calculate ac = (3) (—6), so ac = —18. Now can you think of an 
integer pair whose product is ac = —18 and whose sum is b = —7? For some, 
the pair just pops into their head: 2 and —9. Break up the middle term into a 
sum of like terms using the pair 2 and —9. 



3x 2 -7a;-6 

= 3a; 2 +2a; - 9a; - 6 

= x (3a; + 2) - 3 (3a; + 2) 

= (x-3)(3x + 2) 

Use the FOIL shortcut to check your answer 

F 



(x — 3)(3a; - 



3x 2 





— 7a; = 2x 


-9x. 




Factor by 


grouping 




Factor out 


(3a; + 2) 





/ 


L 


2a; 


- 9x 


- 6 



6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 



429 



Combining like terms, (x — 3)(3a; + 2) = 3a; 2 — 7x — 6, the original trinomial. Our 
solution checks. Note that if you combine the "Outer" and "Inner" products 
mentally, the check goes even faster. 



Answer: (x — 2)(2x — 5) 



□ 



On the other hand, some readers might be saying "Well, the needed ordered 
pair is not popping into my head. Do I have a way of cutting down the work?" 
The answer is "Yes!" As you are listing the ordered pairs whose product equals 
ac, be mindful that you need the ordered pair whose sum is b. If you stumble 
across the needed pair, stop the listing process and "drop" your ordered pair 
in place. 



EXAMPLE 3. Factor: 3x 2 - 33x + 54. 

Solution: Compare 3a; 2 — 33x + 54 with ax 2 + bx + c and note that a = 3, 
b = —33, and c = 54. Calculate ac = (3)(54), so ac = 162. Ouch! That's a big 
number! However, start listing the integer pairs whose product is ac = 162, 
but be mindful that you need an integer pair whose sum is b = —33. 



You Try It! 



Factor: 5x 2 - 35x - 40 



1,162 

2,81 
3,54 
6,27 



-6,-27 



As soon as we wrote down the pair 6 and 27, our mind said "the sum of 6 and 
7 is 33." However, we need the sum to equal b = —33, so we boxed —6 and 
— 27 instead. Next, we break up the middle term into a sum of like terms using 
our circled pair. 



3x -33x - 54 

= 3x 2 -6x- 27a; - 54 
= 3x(x-2)-27(x-2) 
= (3x-27)(x-2) 



— 33a; = — 6x — 27x. 
Factor by grouping. 
Factor out (x — 2). 



Oh-oh! Now we realize we can factor 3 out of each term in the first factor! 
= 3(x-9)(x-2) 

We missed taking out the GCF! Let's try again, only this time let's do 
what we are always supposed to do in the first step: Factor out the GCF. 



3a^ 



33x + 54 = 3(x 2 - llx + 18) 



430 



CHAPTER 6. FACTORING 



Answer: 5 (a; — 8)(x + 1) 



Comparing x 2 — l\x + 18 with ax 2 + bx + c, we see that a = 1, b = — 11, 
and c = 18. We need an integer pair whose product is ac = 18 and whose 
sum is b = — 11. Note that these numbers are considerably smaller than the 
numbers we had to deal with when we forgot to first factor out the GCF. 
Because the numbers are smaller, the integer pair —9 and —2 easily conies to 
mind. Furthermore, because a = 1, we can factor x 2 — 11a; + 18 by simply 
dropping the integer pair —9 and —2 in place. 

3(x 2 - Ux + 18) = 3(x - 9)(x - 2) 
A far simpler solution! 

□ 

In Example 3, we saw how much more difficult we made the problem by 
forgetting to first factor out the greatest common factor (GCF). Let's try not 
to make that mistake again. 

First rule of factoring. The first step in factoring any polynomial is to factor 
out the greatest common factor. 



You Try It! 



Factor: 12a; 4 + 2a; 3 - 30a; 2 



EXAMPLE 4. Factor: 30a; 3 - 21a; 2 - 18a;. 

Solution: Note that the GCF of 30a; 3 , 21a; 2 , and 18a; is 3a;. Factor out this 
GCF. 



— 3a; ■ 7a; 

- 7x - 6) 
c and note that a 



30x^ - 2\x z - 18a; = 3a; ■ lOa;^ - 3a- ■ Ix - 3a; • 6 
= 3a;(10a; 2 

Next, compare 10a: 2 — 7a; — 6 with aa: 2 + bx - 

and c = — 6. Start listing the integer pairs whose product is ac 

mindful that you need an integer pair whose sum is b = — 7. 

1,-60 
2,-30 
3,-20 
4,-15 



10, b = -7, 
—60, but be 



12 



Break up the middle term into a sum of like terms using our circled pair. 



Hence, 30ar 



3a;(10a; 2 - 7x - 6) 

= 3a;(10a; 2 + 5a; - 12a;- 18) 
= 3a;[5a;(2a;+ 1) - 6(2a; + 1)] 
= 3a;(5a;-6)(2a; + l) 

21a; 2 - 18a; = 3a;(5a;-6)(2x 



— 7a; = 5a; — 12a;. 
Factor by grouping. 
Factor out a 2a; + 1 . 



6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 



431 



Check: First, use the FOIL shortcut to multiply the two binomial factors, 
then distribute the monomial factor. 

3x(5x - 6) (2x + 1) = 3x(10a; 2 - 7x - 6) Apply the FOIL shortcut. 
= 30a; 3 - 2 la; 2 - 18a; Distribute the 3a;. 

Because this is the original polynomial, the solution checks. 



Answer: 2x 2 (3x + 5)(2x - 3) 



□ 



Nonlinear Equations Revisited 

Let's use the factoring technique of this chapter to solve some nonlinear equa- 
tions. 



You Try It! 



EXAMPLE 5. Solve the equation 2a; 2 
graphically, then compare your answers. 



13a; — 20 both algebraically and Solve the equation 

5a; 2 = 12a; + 9 both 
algebraically and graphically, 



Solution: Because there is a power of x larger than one, the equation is 
nonlinear. Make one side equal to zero. 

2a; = 13x — 20 Original equation. 

2a; — 13a; + 20 = Make one side zero. 

Compare 2a; 2 — 13a; + 20 with ax 2 + bx + c and note that a = 2, b = — 13 and 
c = 20. We need an integer pair whose product is ac = 40 and whose sum is 
b = —13. The integer pair —5 and —8 comes to mind. Write the middle term 
as a sum of like terms using this pair. 



then compare your answers. 



2a; 2 - 5a; - 8a; + 20 = 

x(2x - 5) - 4(2x - 5) = 

(x-4)(2a;-5) = 



— 13x = — 5x — 8a;. 
Factor by grouping. 
Factor out 2x — 5. 



We have a product that equals zero. Use the zero product property to complete 
the solution. 



Thus, the solutions of 2a; 



a;-4 = 


or 2x — 5 = 


x = 4 


2a; = 5 




5 
X= 2 


2a; 2 = 13a; - 


- 20 are x = 4 and x 



5/2. 



Graphical solution. Load each side of the equation 2x 2 = 13a; — 20 into the 
Y= menu of your graphing calculator, y = 2x 2 in Yl, y = 13a; — 20 in Y2 



432 



CHAPTER 6. FACTORING 



(see Figure 6.18). Select 6:ZStandard from the ZOOM menu to produce the 
image on the left in Figure 6.18. However, even after adjusting the WINDOW 
parameters (Xmin = —10, Xmax = 10, Ymin = —10, and Ymax = 60), the 
image resulting from pushing the GRAPH button (see the image on the right 
in Figure 6.18) does not clearly show the two points of intersection. 




Figure 6.18: Sketch y = 2x 2 and y = 13x - 20. 

Let's switch our strategy and work with the equation 

2x 2 - 13x + 20 = 

instead. Load y = 2x 2 — 13x + 20 into Yl in the Y= menu, then select 
6:ZStandard from the ZOOM menu to produce the image at the right in 
Figure 6.19. 



Plotl PlotE Plots 
WiB2*X A 2-13*X+2 


Wh = 



Figure 6.19: Sketch y = 2x 2 - 13a; + 20. 



To find the solutions of 2a; 2 — 13a; + 20 = 0, we must identify the a;-intercepts of 
the graph in Figure 6.18. Select 2:zero from the CALC menu, then move the 
left- and right=arrows to move the cursor to the left of the first x-intercept. 
Press ENTER to mark the "Left bound," then move the cursor to the right of 
the a;-intercept and press ENTER to mark the "Right bound." Finally, press 
ENTER to use the current position of the cursor for your "Guess." The result 
is shown in the image on the left in Figure 6.20. Repeat the process to find 
the rightmost x-intercept. The result is shown in the image on the right in 
Figure 6.20. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.21). 



6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 



433 





IV 


Z*K* 
K=2.£ 


V=0 





M 


Z*K* 





Figure 6.20: Use 2:zero from the CALC menu to find the x- intercepts. 

• Place your WINDOW parameters at the end of each axis (see Figure 6.21). 

• Label the graph with its equation (see Figure 6.21). 

• Drop dashed vertical lines through each ^-intercept. Shade and label 
the a;-values of the points where the dashed vertical line crosses the x- 
axis. These are the solutions of the equation 2a: 2 — 13a; + 20 = (see 
Figure 6.21). 



10 



-10 




y = 2x 2 - 13a; + 20 



-*- x 



10 



-10 
Figure 6.21: Reporting your graphical solution on your homework. 



Finally, note how the graphical solutions of 2a; 2 — 13a: + 20 = 0, namely x = 2.5 
and x = 4, match the solutions x = 5/2 and x = 4 found using the algebraic 
method. This is solid evidence that both methods of solution are correct. 
However, it doesn't hurt to check the final answers in the original equation, 



434 



CHAPTER 6. FACTORING 



Answer: —3/5, 3 



V 




\ 


V=0 



...A 




\ 

K=3 


V=0 



substituting 5/2 for x and 4 for x. 



and 2aT = 13a; - 20 

2(4) 2 = 13(4) - 20 

2(16) = 13(4) -20 
32 = 52 - 20 



Because the last two statements are true statements, the solutions x = 5/2 and 
x = 4 check in the original equation 2a; 2 = 13a; — 20. 

□ 





2x z 


= 13a; - 20 


2 


<$) 


= 13 (I) " 2 ° 


2 


:d 


= 13(0-20 




25 


65 40 




2 


2 2 



You Try It! 



Solve the equation 



4a; d 



14a; both 



algebraically and graphically, 
then compare your answers. 



EXAMPLE 6. Solve the equation 2x 3 
graphically, then compare your answers. 



28a; both algebraically and 



Solution: Because there is a power of x larger than one, the equation is 
nonlinear. Make one side equal to zero. 

2a; + x = 28a; Original equation. 

2x + x — 28a; = Make one side zero. 

Note that the GCF of 2a; 3 , a; 2 , and 28a; is x. Factor out x. 



x(2x 2 + x - 28) = 



Factor out the GCF. 



Compare 2a; 2 + a;— 28 with aa; 2 + 6a: + c and note that a = 2, b = 1 and c = —28. 
We need an integer pair whose product is ac = —56 and whose sum is b = 1. 
The integer pair —7 and 8 comes to mind. Write the middle term as a sum of 
like terms using this pair. 



a;(2a; 2 - 7x + 8x - 28) = 

a;[a;(2a;-7) + 4(2a;-7)] =0 

a;(a; + 4)(2a;-7) = 



x = — 7x + 8x. 
Factor by grouping. 
Factor out 2a; — 7. 



We have a product of three factors that equals zero. By the zero product 
property, at least one of the factors must equal zero. 







a; + 4 
x 







2a; - 7 = 
2a; = 7 

7 
2 



x 



6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 



435 



Thus, the solutions of 2a; 3 



28a; are x = 0, x = —4, and x = 7/2. 



Graphical solution. Rather than working with 2a; 3 



28x, graphing each 



side separately and finding where the graphs intersect, we will work instead 
28a; = 0, locating where the graph of y = 2a; 3 + x 2 — 28a; 
; 2 — 28a; into Yl in the Y= menu, then 



with 2ar 



crosses the a:-axis. Load y = 2x + : 

select 6:ZStandard from the ZOOM menu to produce the image at the right 

in Figure 6.22. 



Plotl PlotE Plots 
WiB2*X A 3+>T-2-28 

Wh = 



If 



i I 



Figure 6.22: Sketch y = 2x 3 



28a;. 



In the image at the right in Figure 6.22, we saw the graph rise from the bottom 
of the screen, leave the top of the screen, return and leave via the bottom of 
the screen, and then finally return and leave via the top of the screen. Clearly, 
there are at least two turning points to the graph that are not visible in the 
current viewing window. Set the WINDOW settings as shown in the image on 
the left in Figure 6.23, then push the GRAPH button to produce the image on 
the right in Figure 6.23. Note that this window now shows the a;-intercepts as 
well as the turning points of the graph of the polynomial. 



WINDOW 
Xnin=-10 
Xnax=10 
Kscl=l 
Ymin=-100 
Vnax=100 
Yscl=10 




Figure 6.23: Adjusting the view so that the turning points of the polynomial 
are visible. 



To find the solutions of 2a; 



28a; = 0, we must identify the x-intercepts 



of the graph in Figure 6.23. Select 2:zero from the CALC menu, then use the 
left- and right-arrow keys to move the cursor to the left of the first a;- intercept. 
Press ENTER to mark the "Left bound," then move the cursor to the right of 
the a;-intercept and press ENTER to mark the "Right bound." Finally, press 
ENTER to use the current position of the cursor for your "Guess." The result 
is shown in the first image on the left in Figure 6.24. Repeat the process to 
find the remaining a;-intercepts. The results are shown in the next two images 
in Figure 6.24. 



436 



CHAPTER 6. FACTORING 





/ 


f' " 

Z*K* J 

k=-h r 


KJ 







Z*K* J 

k=o r 


v=o 





J 


r ' ' ■ 

Z*K* J 
K=3.E r 


jj6 



Figure 6.24: Use 2:zero from the CALC menu to find the x- intercepts. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.25). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.25) . 

• Label the graph with its equation (see Figure 6.25). 

• Drop dashed vertical lines through each x-intercept. Shade and label 
the x-values of the points where the dashed vertical line crosses the pr- 
axis. These are the solutions of the equation 2a; 3 + x 2 — 28a; = (see 
Figure 6.25). 



y = 2x 3 



28a; 



Answer: -2, 0, 7/4 




Z*K* 
K=-£ 



Z*K0 
K=1.7E 







Z*K0 J 

K=0 I 


V=0 








100' 






10 


-4/ 



-100' 


V /f 3.5 

1 


10 



Figure 6.25: Reporting your graphical solution on your homework. 



x = 0, and x 

using the algebraic method 



3_L/r.2 OV ,- II ,,-,,,.,,1,- ,- /J. 

3.5, match the solutions x = —4, x = 0, and x = 7/2 found 



Finally note how the graphical solutions of 2x A -\-x 2 — 28a; = 0, namely x 



□ 



6.4. FACTORING AX 2 + BX + C WHEN A^l 437 

**.**.**. Exercises -*s •*& >•$ 

In Exercises 1-6, factor the quadratic polynomial. 

1. 6x 2 + 13x - 5 4. 6a; 2 - 23a; + 7 

2. 3a; 2 - 19a; + 20 5. 3a; 2 + 19a; + 28 

3. 4a; 2 -x -3 6. 2a; 2 - 9a; - 18 

In Exercises 7-12, factor the quadratic polynomial. 

7. 12a; 2 - 23a; + 5 10. 4a; 2 + 19a; + 21 

8. 8a; 2 + 22a; + 9 11. 3a; 2 + 4a; - 32 

9. 6a; 2 + 17a; + 7 12. 4a; 2 + x - 14 

In Exercises 13-18, factor the quadratic polynomial. 

13. 3a; 2 + 19a; + 28 16. 4a; 2 - x - 14 

14. 6a; 2 + a;-l 17. 6a; 2 - 11a; - 7 

15. 4a; 2 -21a; + 5 18. 2a; 2 - 17a; + 21 

In Exercises 19-26, factor the trinomial. 

19. 16a; 5 - 36a; 4 + 14a; 3 23. 6a; 4 - 33a; 3 + 42a; 2 

20. 12a; 4 - 20a; 3 + 8a; 2 24. 15a; 3 - 10a; 2 - 105a; 

21. 36a; 4 - 75a; 3 + 21a; 2 25. 16a; 4 - 36a; 3 - 36a; 2 

22. 6a; 4 - 10a; 3 - 24a; 2 26. 40a; 4 - 10a; 3 - 5a; 2 

In Exercises 27-38, use an algebraic technique to solve the given equation. 

27. 4a; 2 = -x + 18 30. 2a; 2 - 20 = -3a; 

28. 2a; 2 = 7x-3 31. 3a; 2 + 30 = 23a; 

29. 3a; 2 + 16 = -14a; 32. 6a; 2 - 7 = -11a; 



438 CHAPTER 6. FACTORING 

33. -7x - 3 = -6a; 2 36. -23a; + 7 = -6a; 2 

34. 13a; -45 = -2a; 2 37. 6x 2 = -25a; + 9 

35. 26a;- 9 = -3a; 2 38. 2.x 2 = 13a; + 45 



In Exercises 39-42, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results 
found using graphing calculator as shown in Example 5. 

39. 2a; 2 - 9a; - 5 = 41. 4a; 2 - 17a; - 15 = 

40. 2a; 2 + x - 28 = 42. 3a; 2 + 14a; - 24 = 



In Exercises 43-46, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results 
found using graphing calculator as shown in Example 6. 

43. 2a; 3 = 3a; 2 + 20a; 45. 10a; 3 + 34a; 2 = 24a; 

44. 2a; 3 = 3x 2 + 35a; 46. 6x 3 + 3a; 2 = 63a; 

$*-$*>£»■ Answers •*$ -*$ •*$ 



1. 


(2a; + 5)(3a;- 1) 


3. 


[x- l)(4a; + 3) 


5. 


^a; + 4)(3a; + 7) 


7. 


^3a;-5)(4a;- 1) 


9. 


^2a; + l)(3a; + 7) 


11. 


^a; + 4)(3a;-8) 


13. 


(a; + 4)(3a; + 7) 


15. 


[x-5)(ix- 1) 



17. 


(3a; -7) (2a; + 1) 


19. 


2a; 3 (2a;- l)(4x- 7) 


21. 


3a; 2 (3a;- l)(4a;- 7) 


23. 


3a; 2 (a;-2)(2a;-7) 


25. 


4a; 2 (x- 3)(4a; + 3) 


27. 


9 

x = 2,-- 


29. 


x = -2, — 
' 3 


31. 


, 5 

x = o, - 

' 3 



1 

33. x = --. 

3 


3 

2 


35. x = -9, 


1 
3 


37.,= !,- 


9 
~2 



6.4. FACTORING AX 2 + BX + C WHEN A^\ 439 

39. x = -1/2,5 
41. x = -3/4,5 
43. x = 0,-5/2, 4 

45. a; = 0,7/2, -5 



440 



CHAPTER 6. FACTORING 



6.5 Factoring Special Forms 

In this section we revisit two special product forms that we learned in Chapter 
5, the first of which was squaring a binomial. 




Perfect Square Trinomials 

To square a binomial such as (a + b) 2 , proceed as follows: 

1. Square the first term: a 2 

2. Multiply the first and second term, then double: 2ab 

3. Square the last term: b 2 



You Try It! 



Expand: (5a + 2b) 2 



Answer: 25a 2 + 20ab + 46 2 



EXAMPLE 1. Expand: (2x + 3y) 2 

Solution: Using the pattern (a + b) 2 = a 2 + 2ab + b 2 , we can expand (2x + 3y) 2 
as follows: 

(2x + 3j/) 2 = (2x) 2 + 2(2x)(3y) + (3y) 2 
= Ax + Qxy + 9y 

Note how we square the first and second terms, then produce the middle term 
of our answer by multiplying the first and second terms and doubling. 

□ 



You Try It! 



Expand: (2s 3 - lif 



EXAMPLE 2. Expand: (3m 2 - 5v 2 ) 2 

Solution: Using the pattern (a—b) 2 = a 2 — 2ab+b 2 , we can expand (3w 2 — 5i> 2 ) 2 
as follows: 

(3u 2 - 5v 2 ) 2 = (3w 2 ) 2 - 2(3u 2 )(5v 2 ) + (5m 2 ) 2 
= 9m 4 - 30m 2 m 2 + 25m 4 



6.5. FACTORING SPECIAL FORMS 



441 



Note that the sign of the middle term is negative this time. The first and last 
terms are still positive because we are squaring. 



Answer: At 



28sH + 49? 



Once you've squared a few binomials, it's time to do all of the work in your 
head, (i) Square the first term; (ii) multiply the first and second term and 
double the result; and (hi) square the second term. 



You Try It! 



EXAMPLE 3. Expand each of the following: 

a) (2y - 3) 2 b) (4a - 36) 2 c) {x 3 + 5) 2 

Solution: Using the pattern (a ± o) 2 = a 2 ± lab + 6 2 , we expand each 
binomially mentally, writing down the answer without any intermediate steps. 

a) (2j/ - 3) 2 = 4y 2 - Yly + 9 

b) (4a - 3b) 2 = 16a 2 - 24a& + 9b 2 

c) (x 3 + 5) 2 = x 6 + 10a; 3 + 25 



Expand: (5z 4 - 3) : 



Answer: 25a 



You Try It! 



EXAMPLE 4. Factor each of the following trinomials: 

a) 4y 2 - 12y + 9 b) 16a 2 - 24a6 + 9b 2 c) x 6 + 10a; 3 + 25 

Solution: Because of the work already done in Example 3, it is a simple task 
to factor each of these trinomials. 

a) Ay 2 - Yly + 9 = (2y - 3) 2 

b) 16a 2 - 24a6 + 96 2 = (4a - 36) 2 

c) x 6 + 10a; 3 + 25 = (a; 3 + 5) 2 



Factor: 25a; 8 - 30.x 4 



Answer: (5x 4 — 3)" 



30a; 4 + 9 



□ 



Now, because factoring is "unmultiplying," it should be a simple matter to 
reverse the process of Example 3. 



□ 



□ 



442 



CHAPTER 6. FACTORING 



Each of the trinomials in Example 4 is an example of a perfect square trinomial. 

Perfect square trinomial. If a trinomial a 2 + 2ab + b 2 is the square of a 
binomial, as in (a + b) 2 , then the trinomial is called a perfect square trinomial. 



You Try It! 



Factor: 16a; 2 + 72a; + 81 



List of Squares 



n 


n 2 








1 


1 


2 


4 


3 


9 


4 


16 


5 


25 


6 


36 


7 


49 


8 


64 


9 


81 


10 


100 


11 


121 


12 


144 


13 


169 


14 


196 


15 


225 


16 


256 


17 


289 


18 


324 


19 


361 


20 


400 


21 


441 


22 


484 


23 


529 


24 


576 


25 


625 



So, how does one recognize a perfect square trinomial? If the first and last 
terms of a trinomial are perfect squares, then you should suspect that you may 
be dealing with a perfect square trinomial. However, you also have to have the 
correct middle term in order to have a perfect square trinomial. 



EXAMPLE 5. Factor each of the following trinomials: 

a) 9a; 2 - 42a; + 49 b) 49a 2 + 70ab + 256 2 c) 4a; 2 - 37a; + 9 

Solution: Note that the first and last terms of each trinomial are perfect 
squares. 

a) In the trinomial 9a; 2 — 42a; + 49, note that (3a;) 2 = 9a; 2 and 7 2 = 49. Hence, 
the first and last terms are perfect squares. Taking the square roots, we 
suspect that 9a; 2 — 42a; + 49 factors as follows: 



9a; 2 



42a; + 49 = (3.x - 7) 2 



However, we must check to see if the middle term is correct. Multiply 3a; 
and 7, then double: 2 (3a;) (7) = 42a; . Thus, the middle term is correct and 
therefore 



9a; 2 



42a; + 49 = (3a; - 7)' 



b) In the trinomial 49a 2 +70a6+256 2 , note that (7a) 2 = 49a 2 and {5b) 2 = 25b 2 . 
Hence, the first and last terms are perfect squares. Taking the square roots, 
we suspect that 49a 2 + 70a6 + 256 2 factors as follows: 

49a 2 + 70a6 + 25o 2 = (7a + 5o) 2 

However, we must check to see if the middle term is correct. Multiply 7a 
and 56, then double: 2(7a)(56) = 70a6. Thus, the middle term is correct 
and therefore 

49a 2 + 70a6 + 256 2 = (7a + 56) 2 . 



c) In the trinomial 4a; 2 - 37a; + 9, note that (2a;) 2 = 4a; 2 and (3) 2 = 9. Hence, 
the first and last terms are perfect squares. Taking the square roots, we 
suspect that 4a; 2 — 37a; + 9 factors as follows: 



4a; 2 - 37a; + 9 = (2a; - 3) 2 



6.5. FACTORING SPECIAL FORMS 



443 



However, we must check to see if the middle term is correct. Multiply 2x 
and 3, then double: 2(2a;)(3) = 12.t. However, this is not the middle term 
of 4a; 2 — 37a; + 9, so this factorization is incorrect! We must find another 
way to factor this trinomial. 

Comparing 4a; 2 — 37a; + 9 with ax 2 + bx + c, we need a pair of integers 
whose product is ac = 36 and whose sum is b = —37. The integer pair —1 
and —36 comes to mind. Replace the middle term as a sum of like terms 
using this ordered pair. 



4a; 2 - 37a; + 9 



4a; 2 - x - 36a; + 9 
a;(4a; - 1) - 9(4a; - 
(a;-9)(4a;- 1) 



—37a; = —x — 36a;. 
1) Factor by grouping. 

Factor out 4a; — 1. 



This example clearly demonstrates how important it is to check the middle 
term. 



Answer: (4a; + 9)^ 



Remember the first rule of factoring! 

The first rule of factoring. The first step to perform in any factoring 
problem is factor out the GCF. 



□ 



You Try It! 



EXAMPLE 6. Factor each of the following trinomials: 



Factor: 



-4a; d 



24a; 2 - 36.x 



a) 2a; 3 y + 12x 2 y 2 + 18xy 3 



b) -4a; 5 + 32a; 4 



64x 3 



Solution: Remember, first factor out the GCF. 



In the trinomial 2x 3 y + 12x 2 y 2 + I8xy 3 , we note that the GCF of 2x 3 y, 
\2x 2 y 2 , and 18xy 3 is 2a;y. We first factor out 2xy. 

2x 3 y + 12x 2 y 2 + 18xy 3 = 2xy{x 2 + Qxy + 9y 2 ) 

We now note that the first and last terms of the resulting trinomial factor 
are perfect squares, so we take their square roots and factors as follows. 



= 2a;y(a; + 3y) 2 

Of course, the last factorization is correct only if the middle term is correct. 
Because 2(x)(3y) = Qxy matches the middle term of a; 2 + 6xy + 9y 2 , we do 
have a perfect square trinomial and our result is correct. 



Answer: — Ax(x + 3)' 



444 CHAPTER 6. FACTORING 



b) In the trinomial -4a; 5 + 32a: 4 - 64a; 3 , we note that the GCF of 4a; 5 , 32.x 4 , 
and 64a: 3 is 4a: 3 . We first factor out 4a; 3 . 

-4a; 5 + 32a; 4 - 64a; 3 = 4a; 3 (-x 2 + 8x - 16) 

However, the first and third terms of —x 2 + 8a; — 16 are negative, and thus 
are not perfect squares. Let's begin again, this time factoring out — 4a; . 

-4a; 5 + 32a; 4 - 64a; 3 = -4a; 3 (x 2 - 8x + 16) 

This time the first and third terms of a; 2 — 8a; + 16 are perfect squares. We 
take their square roots and write: 

= -4a; 3 (a;-4) 2 

Again, this last factorization is correct only if the middle term is correct. 
Because 2 (a;) (4) = 8a;, we do have a perfect square trinomial and our result 
is correct. 



□ 



The Difference of Squares 

The second special product form we learned in Chapter 5 was the difference of 
squares. 



The difference of squares. Here is the difference of squares rule. 

(a + b)(a-b) = a 2 - b 2 



If you are multiplying two binomials which have the exact same terms in 
the "First" positions and the exact same terms in the "Last" positions, but 
one set is separated by a plus sign while the other set is separated by a minus 
sign, then multiply as follows: 

1. Square the first term: a 2 

2. Square the second term: b 2 

3. Place a minus sign between the two squares. 



6.5. FACTORING SPECIAL FORMS 445 



You Try It! 



EXAMPLE 7. Expand each of the following: Expand: (4c - 3y)(4a; + 3y) 

a) (3a; + 5)(3cc - 5) b) (a 3 - 26 3 )(a 3 + 2b 3 ) 

Solution: We apply the difference of squares pattern to expand each of the 
given problems. 

a) In (3x + 5) (3a; — 5), we have the exact same terms in the "First" and 
"Last" positions, with the first set separated by a plus sign and the second 
set separated by a minus sign. 

a) Square the first term: (3a;) 2 = 9x 2 

b) Square the second term: 5 2 = 25 

c) Place a minus sign between the two squares. 

Hence: 

(3a; + 5)(3x-5) = 9a; 2 - 25 

b) In (a 3 — 26 3 )(a 3 + 2b 3 ), we have the exact same terms in the "First" and 
"Last" positions, with the first set separated by a minus sign and the second 
set separated by a plus sign. 

a) Square the first term: (a 3 ) 2 = a 6 

b) Square the second term: (26 3 ) 2 = 4& 6 

c) Place a minus sign between the two squares. 

Hence: 

(a 3 -26 3 )(a 3 + 26 3 ) = a 6 -46 6 

Answer: 16a; 2 — 9j/ 2 

□ 

Because factoring is "unmultiplying," is should be a simple matter to reverse 
the process of Example 7. 



You Try It! 



EXAMPLE 8. Factor each of the following: Factor: 81a; 2 - 49 

a) 9a; 2 - 25 b) a 6 - 46 6 

Solution: Because of the work already done in Example 7, it is a simple 
matter to factor (or "unmultiply" ) each of these problems. 

a) 9a; 2 -25= (3a; + 5) (3a; - 5) 

b) a 6 -4b 6 = {a 3 -2b 3 ) (a 3 + 2b 3 ) 



446 



CHAPTER 6. FACTORING 



Answer: (9x + 7)(9x - 7) 



In each case, note how we took the square roots of each term, then separated 
one set with a plus sign and the other with a minus sign. Because of the 
commutative property of multiplication, it does not matter which one you 
make plus and which one you make minus. 

□ 



Always remember the first rule of factoring. 

The first rule of factoring. The first step to perform in any factoring 
problem is factor out the GCF. 



You Try It! 



Factor: 4x — 16.x 



EXAMPLE 9. Factor: x 3 - 9x 

Solution: In x 3 — 9x, the GCF of a; 3 and 9x is x. Factor out x. 

„3 n~. ™/~,2 



X 



9x = x(x z - 9) 



Answer: Ax 2 (x + 2) (x - 2) 



Note that x 2 — 9 is now the difference of two perfect squares. Take the square 
roots of x 2 and 9, which are x and 3, then separate one set with a plus sign 
and the other set with a minus sign. 

= x(x + 3)(x — 3) 



□ 



You Try It! 



Factor: x — 81 



Factoring Completely 

Sometimes after one pass at factoring, factors remain that can be factored 
further. You must continue to factor in this case. 



EXAMPLE 10. Factor: x 4 



16 



Solution: In 



16, we have the difference of two squares: (x 2 ) z 



and 



4 2 = 16. First, we take the square roots, then separate one set with a plus sign 
and the other set with a minus sign. 

x 4 - 16= (x 2 +4)(x 2 -4) 



6.5. FACTORING SPECIAL FORMS 



447 



Note that x 2 +4 is the sum of two squares and does not factor further. However, 
x 2 — 4 is the difference of two squares. Take the square roots, x and 2, then 
separate one set with a plus sign and the other set with a minus sign. 



(a; 2 +4)(a; + 2)(a--2) 



Done. We cannot factor further. 



Answer: 

(cc 2 +9)(a; + 3)(a;-3) 



□ 



Nonlinear Equations Revisited 

Remember, if an equation is nonlinear, the first step is to make one side equal 
to zero by moving all terms to one side of the equation. Once you've completed 
this important first step, factor and apply the zero product property to find 
the solutions. 



EXAMPLE 11. Solve for 



25a; 2 



169 



Solution: Make one side equal to zero, factor, then apply the zero product 
property. 



2hx z 



25x = 169 Original equation. 

- 169 = Subtract 169 from both sides. 



You Try It! 



Solve for 



16x 2 



121 



Note that we have two perfect squares separated by a minus sign. This is the 
difference of squares pattern. Take the square roots, making one term plus and 
one term minus. 



(5x + 13)(5x- 13) = 



Use difference of squares to factor. 



Use the zero product property to complete the solution, setting each factor 
equal to zero and solving the resulting equations. 



5z + 13 = 



5x- 13 



13 




13 

~5~ 



Hence, the solutions of 25a; 2 = 169 are x = —13/5 and x = 13/5. We encourage 
readers to check each of these solutions. 



Answer: -11/4, 11/4 



□ 



448 



CHAPTER 6. FACTORING 



You Try It! 



Solve for x: 
25a; 2 = 80a; 



64 



EXAMPLE 12. Solve for x: 



49a; 2 + 81 = 126a; 



Solution: Make one side equal to zero, factor, then apply the zero product 
property. 



One can also argue that the 
only number whose square 
is zero is the number zero. 
Hence, one can go directly 
from 



(7a -9) 5 







to 



7x - 9 = 0. 



Hence, the only solution of 
49a; 2 + 81 = 126a; is x = 9/7. 



49x 2 



n 



<±9x z - 126a; + 81 



126a; Original equation. 

Subtract 126a; from both sides. 



Note that the first and last terms of the trinomial are perfect squares. Hence, 
it make sense to try and factor as a perfect square trinomial, taking the square 
roots of the first and last terms. 



(7a;-9) 2 = 



Factor as a perfect square trinomial. 



Of course, be sure to check the middle term. Because — 2(7a;)(9) = —126a;, the 
middle term is correct. Because (7a; — 9) 2 = (7x — 9) (7a; — 9), we can use the 
zero product property to set each factor equal to zero and solve the resulting 
equations. 



Answer: 8/5 



7a; - 9 = 
9 

X= 7 



7x - 9 = 

9 

X= 7 



Hence, the only solution of 49a; 2 + 81 = 126a; is x = 9/7. We encourage readers 
to check this solution. 

□ 



You Try It! 



Solve for 
5a; 3 + 36 



180a; 



EXAMPLE 13. Solve for 



2a; 3 + 3a; 2 = 50a; + 75 



Solution: Make one side equal to zero, factor, then apply the zero product 
property. 



2a; + 3a; = 50a; + 75 Original equation. 
Make one side zero. 



2x 3 + 3a; 2 - 50a; - 75 = 



This is a four-term expression, so we try factoring by grouping. Factor a; 2 out 
of the first two terms, and —25 out of the second two terms. 



a; 2 (2a; + 3)-25(2a; + 3) = 
(a; 2 -25)(2x + 3) = 



Factor by grouping 
Factor out 2a; + 3. 



6.5. FACTORING SPECIAL FORMS 



449 



Complete the factorization by using the difference of squares to factor x 2 — 25. 

(x + 5)(x — 5) (2x + 3) = Use difference of squares to factor. 

Finally, use the zero product property. Set each factor equal to zero and solve 
for x. 



x + 5 = 
x = —5 



or 



x-5 = 

x = 5 



or 



2x 







Hence, the solutions of 2x 3 + 3x 2 = 50a; + 75 are x = —5, x = 5, and x 
We encourage readers to check each of these solutions. 



-3/2. 



Answer: —6, 6, 1/5 



Let's solve another nonlinear equation, matching the algebraic and graphi- 
cal solutions. 



□ 



You Try It! 



EXAMPLE 14. Solve the equation x 3 = 4a;, both algebraically and graphi- Solve the equation a; 3 = 16a; 



cally, then compare your answers. 

Solution: Note that we have a power of x larger than one, so the equation 
a; 3 = 4a; is nonlinear. Make one side zero and factor. 



4a; 



a; 3 - 4a; = 

x(x 2 - 4) = 

x{x + 2){x-2) = 



Original equation. 

Nonlinear. Make one side zero. 

Factor out GCF. 

Apply difference of squares. 



both algebraically and 
graphically, then compare 
your answers. 



a; + 2 = 


or x - 2 = 


x = -2 


x = 2 


4x are x = 0, x 


= —2, and x = 2. 



Note that we now have a product of three factors that equals zero. The zero 
product property says that at least one of these factors must equal zero. 







Hence, the solutions of a; 3 

Graphical solution. Load y = x 3 and y = 4x into Yl and Y2 in the 
Y= menu of your calculator. Select 6:ZStandard from the ZOOM menu to 
produce the graph in Figure 6.26. 

Although the image in Figure 6.26 shows all three points of intersection, 
adjusting the WINDOW parameters as shown in Figure 6.27, then pressing 
the GRAPH button will produce a nicer view of the points of intersection, as 
shown in the figure on the right in Figures 6.27. 



450 



CHAPTER 6. FACTORING 



Plotl PlotE 


PlOtS 


WiBX A 3 




\Y*B4*IK 




Wj=I 




^Vh= 




We = 




N.Vfi = 




\V? = 





Figure 6.26: Sketching y 



WINDOW 


Xmin= _ 5 


Xmax=5 


Xscl=l 


Vnin=-15 


Vmax=15 


Yscl=5 


4-Xres=i 











= a; 3 and y = Ax. 











Figure 6.27: Adjusting the viewing window. 



Use the 5:intersect tool from the CALC menu to find the three points of 
intersection. Press the ENTER key in response to "First curve," then press 
ENTER again in response to "Second curve," then use the left-arrow key to 
move your cursor close to the leftmost point of intersection and press ENTER 
in response to "Guess." The result is shown in the first image on the left in 
Figure 6.28. Repeat the process to find the remaining points of intersection. 
The results are shown in the last two images in Figure 6.28. 







Intersection 
K=-£ / 


v=-H 







Intersection 
K=0' / 


Y=0 





i y 


Intersection 
K=2' i 


V=B 



Figure 6.28: Finding the points of intersection. 



Thus, the graphical solutions are x = —2, x = 0, and x = 2. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 6.29). 

• Place your WINDOW parameters at the end of each axis (see Figure 6.29). 

• Label the graph with its equation (see Figure 6.29). 



6.5. FACTORING SPECIAL FORMS 



451 



• Drop dashed vertical lines through each x-intercept. Shade and label the 
a;- values of the points where the dashed vertical line crosses the a;- axis. 
These are the solutions of the equation x 3 = 4x (see Figure 6.29). 




Figure 6.29: Reporting your graphical solution on your homework. 

Finally, note that the graphical solutions x = —2, x = 0, and x = 2 match our 
algebraic solutions exactly. 



Answer: —4, 0, 4 







Intersection 
K=-4 


V="fiH 







Intersection 
K=0 


Y=0 





^ 


Intersection 


V=fiH 



□ 



452 CHAPTER 6. FACTORING 



f> it, t* Exercises ■** •** ■** 

In Exercises 1-8, expand each of the given expressions. 

1. (8r - 3t) 2 5. (s 3 - 9) 2 

2. (6a + c) 2 6. (w 3 + 7) 2 

3. (4a + 76) 2 7. (s 2 +6i 2 ) 2 

4. (4s + t) 2 8. (7u 2 -2w 2 ) 2 

In Exercises 9-28, factor each of the given expressions. 

9. 25s 2 + 60s* + 36£ 2 19. 49r 6 + 112r 3 + 64 

10. 9m 2 + 24uv + 16v 2 20. a e - 16a 3 + 64 

11. 36m 2 - QOvw + 25w 2 21. 5s 3 £ - 20s 2 f 2 + 20st 3 

12. 496 2 - 426c + 9c 2 22. 12r 3 t - 12r 2 i 2 + 3rt 3 

13. a 4 + 18a 2 6 2 + 816 4 23. 8a 3 c + 8a 2 c 2 + 2ac 3 

14. 64m 4 - U4u 2 w 2 + 81w 4 24. 18a; 3 z - 60a; 2 z 2 + 50xz 3 

15. 49s 4 - 28s 2 i 2 + At 4 25. -486 3 + 1206 2 - 756 

16. 4a 4 - 12a 2 c 2 + 9c 4 26. -45c 3 + 120c 2 - 80c 

17. 496 6 - 112b 3 + 64 27. -5m 5 - 30u 4 - 45m 3 

18. 25ir 6 - lOz 3 + 1 28. -12z 5 - 36z 4 - 27z 3 

In Exercises 29-36, expand each of the given expressions. 

29. (21c + 16)(21c - 16) 33. (3j/ 4 + 23z 4 )(3y 4 - 23z 4 ) 

30. (19i + 7)(19t - 7) 34. (5x 3 + z 3 ){5x 3 - z 3 ) 

31. (5a; + 19z)(5a: - 19z) 35. (8r 5 + 19s 5 )(8r 5 - 19s 5 ) 

32. (11m + 5w)(11m - 5w) 36. (3m 3 + 16w 3 )(3m 3 - 16w 3 ) 



6.5. FACTORING SPECIAL FORMS 453 

In Exercises 37-60, factor each of the given expressions. 

37. 361a; 2 - 529 49. 72y 5 - 242y 3 

38. 96 2 - 25 50. 75y 5 - U7y 3 

39. 16v 2 - 169 51. 1444a 3 o - 324a6 3 

40. 81r 2 - 169 52. 126 3 c - 18756c 3 

41. 169a; 2 - 576y 2 53. 576a; 3 z - 1156o;;z 3 

42. 100y 2 - 81z 2 54. 192w 3 v - 507uv 3 

43. 529r 2 - 289s 2 55. 576i 4 - 4£ 2 

44. 49a 2 - 1446 2 56. 4z 5 - 256z 3 

45. 49r 6 - 256£ 6 57. 81a; 4 - 256 

46. 361a; 10 - 484z 10 58. 81a; 4 - 1 

47. 36u 10 - 25w 10 59. 81a; 4 - 16 

48. a 6 - 81c 6 60. a; 4 - 1 

In Exercises 61-68, factor each of the given expressions completely. 

61. z 3 + z 2 -9z-9 

62. 3m 3 + u 2 -48m- 16 

63. a; 3 - 2a; 2 y - xy 2 + 2y 3 

64. a; 3 + 2a; 2 z - 4a;z 2 - 8z 3 

In Exercises 69-80, solve each of the given equations for x. 

69. 2a; 3 + 7a; 2 = 72a; + 252 75. 16a; 2 = 169 

70. 2a; 3 + 7a; 2 = 32a; +112 76. 289a; 2 = 4 

71. a; 3 + 5a; 2 = 64a; + 320 77. 9a; 2 = 25 

72. a; 3 + 4a; 2 = 49a; + 196 78. 144a; 2 = 121 

73. 144a; 2 + 121 = 264x 79. 256a; 2 + 361 = -608a; 

74. 361a; 2 + 529 = 874a; 80. 16a; 2 + 289 = -136a; 



65. 


r 3 - 3r 2 t - 25rt 2 + 75t 3 


66. 


26 3 - 36 2 c - 506c 2 + 75c 3 


67. 


2a; 3 + a; 2 - 32a; - 16 


68. 


r 3 - 2r 2 - r + 2 



454 CHAPTER 6. FACTORING 

In Exercises 81-84, perform each of the following tasks: 

i) Use a strictly algebraic technique to solve the given equation. 

ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the 
results found using graphing calculator as shown in Example 12. 

81. x 3 = x 83. Ax 3 = x 

82. x 3 = 9x 84. 9x 3 = x 

&»■ j*- £*■ Answers •*$ •** *** 

1. 64r 2 -48rt + 9i 2 35. 64r 10 - 361s 10 

3. 16a 2 + 56a6 + 496 2 37. (19a: + 23)(19x - 23) 

5. s 6 - 18s 3 + 81 39. (4w + 13)(4v-13) 

7. s 4 + 12s 2 i 2 + 36i 4 41. 169x 2 - 576y 2 

9. (5s + 6i) 2 43. 529r2 " 289s 2 

11. (6t, - 5u,) 2 45. (7r 3 + 16i 3 )(7r 3 - 16£ 3 ) 

13. (a 2 + 96 2 ) 2 47. (6m 5 + 5w 5 )(6m 5 - 5u> 5 ) 

15. (7s 2 - 2i 2 ) 2 49 - 2y 3 (6y + H)(6» - 11) 

17 (7fe3_ 8 )2 51. 4a6(19a + 96)(19a-96) 

19. (7r 3 + 8) 2 53- 4a:2; ( 12a; + 17z)(12a: - 17z) 

21.5st(s-2tr 55.4t 2 (m+l)(m-l) 

23.2 ac( 2a + C ) 2 57. (9x 2 + 16)(3x + 4)0,-4) 



25. -36(46-5) 



2 



29. 441c 2 - 256 
31. 25ir 2 -361z 2 
33. 9y 8 - 529z 8 



59. (9x 2 + 4)(3x + 2)(3x - 2) 



27.-5.3 (u + 3 ) 2 81.(* + 3)(z-3)(z + l) 



63. (x + j/)(x-y)(x-2j/) 
65. (r + U)(r - 5t)(r - 3t) 
67. (x + 4)(x-4)(2x+l) 



69. x = —6, 6, 



71. x = -8, 8, -5 



6.5. FACTORING SPECIAL FORMS 455 

7 



73. x = 


11 
12 


75. z = 


13 13 

~T'T 



77. 


a; = 


5 5 
~3' 3 


79. 


CE = 


19 
~16 


81. 


a; = 


0,-1,1 



83. x = 0,-1/2,1/2 



456 CHAPTER 6. FACTORING 

6.6 Factoring Strategy 

When you are concentrating on factoring problems of a single type, after doing a 
few you tend to get into a rhythm, and the remainder of the exercises, because 
they are similar, seem to flow. However, when you encounter a mixture of 
factoring problems of different types, progress is harder. The goal of this 
section is to set up a strategy to follow when attacking a general factoring 
problem. 

If it hasn't already been done, it is helpful to arrange the terms of the given 
polynomial in some sort of order (descending or ascending). Then you want to 
apply the following guidelines. 

Factoring Strategy. These steps should be followed in the order that they 
appear. 

1. Factor out the greatest common factor (GCF). 

2. Look for a special form. 

a) If you have two perfect squares separated by a minus sign, use the 
difference of squares pattern to factor: 

a 2 -b 2 = {a + b)(a-b) 

b) If you have a trinomial whose first and last terms are perfect squares, 
you should suspect that you have a perfect square trinomial. Take 
the square roots of the first and last terms and factor as follows. 

a 2 + 2ab + b 2 = (a + b) 2 

Be sure to check that the middle term is correct. 

3. If you have a trinomial of the form ax 2 + bx + c, use the ac- method to 
factor. 

4. If you have a four-term expression, try to factor by grouping. 



Once you've applied the above strategy to the given polynomial, it is quite 
possible that one of your resulting factors will factor further. Thus, we have 
the following rule. 

Factor completely. The factoring process is not complete until none of your 
remaining factors can be factored further. This is the meaning of the phrase, 
"factor completely." 



6.6. FACTORING STRATEGY 457 

Finally, a very good word of advice. 

Check your factoring by multiplying. Once you've factored the given 
polynomial completely, it is a very good practice to check your result. If you 
multiply to find the product of your factors, and get the original given polyno- 
mial as a result, then you know your factorization is correct. 

It's a bit more work to check your factorization, but it's worth the effort. It 
helps to eliminate errors and also helps to build a better understanding of the 
factoring process. Remember, factoring is "unmultiplying," so the more you 
multiply, the better you get at factoring. 

Let's see what can happen when you don't check your factorization! 



Warning! 


! The following solution 


is incorrect! 


Factor: 


2a; 4 + 8a; 2 . 




Solution: 


Factor out the GCF. 






2a; 4 + 8x 2 -■ 


= 2x 2 (x 2 +4) 
= 2x 2 {x + 2) 2 



Note that this student did not bother to check his factorization. Let's do that 
for him now. 

Check: Multiply to check. Remember, when squaring a binomial, there is a 
middle term. 

2x 2 {x + 2) 2 = 2x 2 (x 2 + 4x + 4) 
= 2a; 4 + 8a; 3 + 8a; 2 

This is not the same as the original polynomial 2a; 4 + 8a; 2 , so the student's 
factorization is incorrect. Had the student performed this check, he might 
have caught his error, provided of course, that he multiplies correctly during 
the check. 

The correct factorization follows. 

2a; 4 + 8a; 2 = 2a; 2 (a; 2 +4) 

The sum of squares does not factor, so we are finished. 
Check: Multiply to check. 

2a; 2 (a; 2 +4) = 2a; 4 + 8a; 2 

This is the same as the original polynomial 2a; 4 + 8a; 2 , so this factorization is 
correct. 



458 



CHAPTER 6. FACTORING 



You Try It! 



Factor completely: 
-4a; 7 + 64a; 3 



Answer: 

-4x 3 (a; 2 + 4)(x + 2){x - 2) 



EXAMPLE 1. Factor completely: -3a; 6 



3a; 2 



Solution: The first rule of factoring is "Factor out the GCF." The GCF of 
— 3x 6 and 3a: 2 is 3a; 2 , so we could factor out 3a; 2 . 

-3a; 6 + 3a; 2 = 3a: 2 (-a: 4 + 1) 

This is perfectly valid, but we don't like the fact that the second factor starts 
with —x. Let's factor out —3a; 2 instead. 



-3a; 6 + 3a; 2 



3a; 2 (a; 4 -l) 



The second factor is the difference of two squares. Take the square roots, 
separating one pair with a plus sign, one pair with a minus sign. 

= -3x 2 (a; 2 + l)(a; 2 - 1) 

The sum of squares does not factor. But the last factor is the difference of two 
squares. Take the square roots, separating one pair with a plus sign, one pair 
with a minus sign. 

= -3x 2 (a; 2 + l)(a; + l)(a;- 1) 

Check: Multiply to check the result. 

-3a; 2 (a; 2 + l)(x + l)(x - 1) = -3a; 2 (a; 2 + l)(a; 2 - 1) 

= -3a: 2 (a: 4 -l) 



-3a; 6 + 3a; 2 



The factorization checks. 



□ 



You Try It! 



Factor completely: 
3a 2 6 4 + 12a 4 & 2 - 12a 3 £ 



EXAMPLE 2. Factor completely: a: 3 y + 9a;y 3 + 6a; 2 y 2 

Solution: The first rule of factoring is "Factor out the GCF." The GCF of 
x 3 y, 9a:?/ 3 , and 6x 2 y 2 is xy, so we factor out xy. 

x y + 9xy + 6x y = xy(x +9y + 6xy) 
Let's order that second factor in descending powers of x. 

= xy(x 2 + 6xy + 9y 2 ) 



6.6. FACTORING STRATEGY 



459 



The first and last terms of the trinomial factor are perfect squares. We suspect 
we have a perfect square trinomial, so we take the square roots of the first and 
last terms, check the middle term, and write: 



xy(x + 3j/) 5 



Thus, x 3 y + 9xy 3 + 6x 2 y 2 = xy(x + 3y) 2 . 
Check: Multiply to check the result. 



xy(x + 3y) 2 = xy(x 2 + 6xy + 9y 2 ) 



9xy 3 



x 3 y + 6x 2 y 2 



Except for the order, this result is the same as the given polynomial. The 
factorization checks. 



Answer: 3a 2 6 2 (2a 



a 



EXAMPLE 3. Factor completely: 2x 3 - 48a; + 20a; 2 

Solution: In the last example, we recognized a need to rearrange our terms 
after we pulled out the GCF. This time, let's arrange our terms in descending 
powers of x right away. 



2a; 3 - 48a; + 20a; 2 



2x J 



2Qx z - 48a; 



You Try It! 



Factor completely: 
27a; 3 - 3a; 4 - 60a; 2 



Now, let's factor out the GCF. 



2a;(a; 2 + 10a; - 24) 



The last term of the trinomial factor is not a perfect square. Let's move to the 
ac-method to factor. The integer pair —2, 12 has a product equal to ac = —24 
and a sum equal to b = 10. Because the coefficient of a; 2 is one, this is a "drop 
in place" situation. We drop our pair in place and write: 



Thus, 2x 3 



= 2x(x- 2)(x + 12) 
48a; + 20a; 2 = 2a:(a; - 2)(x + 12). 



Check: Multiply to check the result. We use the FOIL method shortcut and 
mental calculations to speed things up. 

2a;(a; - 2)(a; + 12) = 2a;(a; 2 + 10a; - 24) 
= 2a; 3 + 20a; 2 - 48a; 

Except for the order, this result is the same as the given polynomial. The 
factorization checks. 



Answer: — 3a; 2 (a; — 4)(a; — 5) 



□ 



460 



CHAPTER 6. FACTORING 



You Try It! 



Factor completely: 
8a; 2 + 14xy - 15y 2 



EXAMPLE 4. Factor completely: 2a 2 - 13a6 



246 2 



Solution: There is no common factor we can factor out. We have a trinomial, 
but the first and last terms are not perfect squares, so let's apply the ac-method. 
Ignoring the variables for a moment, we need an integer pair whose product is 
ac = —48 and whose sum is —13. The integer pair 3, —16 comes to mind (if 
nothing comes to mind, start listing integer pairs). Break up the middle term 
into a sum of like terms using the integer pair 3, —16, then factor by grouping. 

2a 2 - 13a6 - 246 2 = 2a 2 + 3a6 - 16a6 - 246 2 
= a(2a + 36) - 86(2a + 36) 
= (a-86)(2a + 36) 

Thus, 2a 2 - 13a6 - 246 2 = (a - 86)(2a + 36). 

Check: Multiply to check the result. We use the FOIL method shortcut and 
mental calculations to speed things up. 



(a - 86)(2a + 36) = 2a 2 - 13a6 - 246 2 
Answer: (2x + 5y)(Ax — 3y) This result is the same as the given polynomial. The factorization checks. 



□ 



You Try It! 



Factor completely: 
36a; 3 + 60a; 2 + 9a; 



EXAMPLE 5. Factor completely: 30a; 4 + 38a; 3 - 20x 2 

Solution: The first step is to factor out the GCF, which in this case is 2a; 2 . 

30a; 4 + 38a; 3 - 20a; 2 = 2a; 2 (15a; 2 + 19a; - 10) 

The first and last terms of the trinomial factor are not perfect squares, so let's 
move again to the ac-method. Comparing 15a; 2 + 19a; — 10 with ax 2 +bx+c, note 
that ac = (15) (—10) = —150. We need an integer pair whose product is —150 
and whose sum is 19. The integer pair —6 and 25 satisfies these requirements. 
Because a/1, this is not a "drop in place" situation, so we need to break up 
the middle term as a sum of like terms using the pair —6 and 25. 



= 2a; 2 (15a; 2 - 6a; + 25a; - 10) 

Factor by grouping. Factor 3a; out of the first two terms and 5 out of the third 
and fourth terms. 



2a; 2 (3a;(5a;-2) + 5(5a;-2)) 



6.6. FACTORING STRATEGY 



461 



Finally, factor out the common factor 5a; — 2. 

= 2a; 2 (3a; + 5)(5a;-2) 

Thus, 30a: 4 + 38a; 3 - 20a: 2 = 2a: 2 (3a; + 5) (5a; - 2). 

Check: Multiply to check the result. Use the FOIL method to first multiply 
the binomials. 

2a; 2 (3a; + 5) (5a; - 2) = 2a; 2 (15a; 2 + 19a; - 10) 

Distribute the 2a; 2 . 

= 30a; 4 + 38a; 3 - 20a; 2 
This result is the same as the given polynomial. The factorization checks. Answer: 3a;(6a: + l)(2x + 3) 



□ 



You Try It! 



EXAMPLE 6. Factor completely: 8a; 5 + 10a; 4 - 72a; 3 - 90a; 2 
Solution: Each of the terms is divisible by 3a; 3 . Factor out 3a; 3 . 

15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 = 3a; 3 [5a; 3 - 11a; 2 - 80a; + 176] 

The second factor is a four-term expression. Factor by grouping. 

= 3a; 3 [a: 2 (5a; - 11) - 16 (5a; - 11)] 
= 3a; 3 (a; 2 - 16) (5s- 11) 

The factor a; 2 — 16 is a difference of two squares. Take the square roots, separate 
one pair with a plus, one pair with a minus. 

= 3a; 3 (a; + 4)(a;-4)(5a;- 11) 

Thus, 15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 = 3x 3 (a; + 4) (a; - 4) (5a; - 11). 
Check: Multiply to check the result. 

3a; 3 (a; + 4) (a; - 4) (5a; - 11) = 3x 3 (a; 2 - 16) (5a; - 11) 

= 3a; 3 (5a; 3 - 11a; 2 - 80a; + 176 
= 15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 

This result is the same as the given polynomial. The factorization checks. 



Factor completely: 
15a; 6 - 33a; 5 - 240a; 4 



528a; 3 



Answer: 

2a; 2 (a;-3)(a; + 3)(4a; + 5) 



□ 



462 



CHAPTER 6. FACTORING 



Using the Calculator to Assist the ac-Method 

When using the ac-method to factor ax 2 + bx + c and ac is a very large number, 
then it can be difficult to find a pair whose product is ac and whose sum in b. 
For example, consider the trinomial: 

12y 2 - lly-36 

We need an integer pair whose product is ac = —432 and whose sum is 6 = —11. 
We begin listing integer pair possibilities, but the process quickly becomes 
daunting. 



432 
216 



Note that the numbers in the second column are found by dividing ac = —432 
by the number in the first column. We'll now use this fact and the TABLE 
feature on our calculator to pursue the desired integer pair. 

1. Enter the expression -432/X into Yl in the Y= menu (see the first image 
in Figure 6.30). 

2. Above the WINDOW button you'll see TBLSET. Use the 2nd key, then 
press the WINDOW button to access the menu shown in the second 
image of Figure 6.30. Set TblStart=l, ATbl=l, then highlight AUTO 
for both the independent and dependent variables. 

3. Above the GRAPH button you'll see TABLE. Use the 2nd key, then 
press the GRAPH button to access the table shown in the third image 
in Figure 6.30. Use the up- and down-arrow keys to scroll through the 
contents of the table. Note that you can ignore most of the pairs, be- 
cause they are not both integers. Pay attention only when they are both 
integers. In this case, remember that you are searching for a pair whose 
sum is b = —11. Note that the pair 16, —27 shown in the third image of 
Figure 6.30 is the pair we seek. 



Ploti 


PlOtE 


plots 


WiB- 


-432/X 


-:Vi=U 




Wj = 






^Vh= 






We = 






^Vi= 






\V? = 







TABLE SETUP 




TblStart=l 




^Tbl=l 






Indpnt: 


grog 


Rsk 


Depend: 


flsk 



X 


V1 




Ff^^B 


-39.27 




12 


"3fi 




13 


-33.23 




1H 


-30.B6 




IE 


-2B.B 




IS 


-27 




1? 


-25.11 




X=ll 



Figure 6.30: Using the TABLE feature to assist the ac-method. 



6.6. FACTORING STRATEGY 463 

Now we can break the middle term of I2y 2 — lly — 36 into a sum of like 
terms using the ordered pair 16, — 27, then factor by grouping. 



• 2 -iiy- 


-36 = 


= 12y 2 


+ 162/- 


-27y- 


-36 






= 4y(3y + 4)- 


-9(32/ 


+ 4) 






= (4y- 


- 9)(3y 


+ 4) 





Check: Use the FOIL method shortcut and mental calculations to multiply. 

(42/-9)(32/ + 4) = 122/ 2 -ll2/-36 
The factorization checks. 



3. 


3x 7 z b - 363x 5 ;j 5 


4. 


5r 5 s 2 - 80r 3 s 2 


5. 


2w 7 - 162u 5 


6. 


405a; 4 - 320a; 2 



464 CHAPTER 6. FACTORING 



f> t* t* Exercises ■** •** ■** 

In Exercises 1-12, factor each of the given polynomials completely. 

1. 484y 4 z 2 - U4y 2 z 4 7. 3t> 8 - 1875w 4 

2. 72s 4 i 4 - 242s 2 t 6 8. 3a 9 - 48a 5 

9. 3x 6 - 300a; 4 

10. 2y 5 - 18y 3 

11. 1250u 7 w 3 - 2w 3 w 7 

12. 48y 8 z 4 - 3y 4 z 8 

In Exercises 13-24, factor each of the given polynomials completely. 

13. 75a 6 - 210a 5 + 147a 4 19. 2^ 4 - 4z 3 + 2z 2 

14. 245v 7 - 560u 6 + 320w 5 20. 2m 6 - 40w 5 + 200m 4 

15. 180a 5 6 3 + 540a 4 6 4 + 405a 3 6 5 21. 324a; 4 + 360a; 3 + 100x 2 

16. 192uV + 432uV + 243u 4 w 6 22. 986 4 + 846 3 + 186 2 

17. 26 5 + 46 4 + 26 3 23. 756 4 c 5 - 2406 3 c 6 + 1926 2 c 7 

18. 3v 6 + 30v 5 + 75u 4 24. 162a 5 c 4 - 180a 4 c 5 + 50a 3 c 6 

In Exercises 25-36, factor each of the given polynomials completely. 

25. 5a 5 + 5a 4 - 210a 3 31. 4a 6 + 64a 5 + 252a 4 

26. 3y 5 - 9y 4 - 12y 3 32. 4x 4 + 64a; 3 + 252a; 2 

27. 3y 6 - 39y 5 + 120w 4 33. 3z 4 + 33z 3 + 84z 2 

28. 3y 7 - 27y 6 + 42y 5 34. 5a 6 + 65a 5 + 180a 4 

29. 3z 4 + I2z 3 - I3hz 2 35. bz 7 - 75z 6 + 270z 5 

30. 5a 4 - 40a 3 - 45a 2 36. 3y 4 - 27y 3 + 24y 2 



6.6. FACTORING STRATEGY 465 

In Exercises 37-48, factor each of the given polynomials completely. 

37. 46 3 - 22b 2 + 306 43. 12a; 3 + 9a; 2 - 30a; 

38. 46 6 - 22b 5 + 30b 4 44. 6v 4 + 2v 3 - 20v 2 

39. 2u 4 w 5 - 3u 3 w 6 - 20u 2 w 7 45. 8m 6 + 34w 5 + 30u 4 

40. 12x 5 z 2 + 9x 4 z 3 - 30x 3 z 4 46. 4a 4 + 29a 3 + 30a 2 

41. 12x 4 y 5 + 50x 3 y 6 + 50x 2 y 7 47. 12a 4 c 4 - 35a 3 c 5 + 25a 2 c 6 

42. 24s 4 t 3 + 62s 3 i 4 + 40s 2 i 5 48. 18x 6 z 5 - 39x 5 z 6 + 18x 4 z 7 

In Exercises 49-56, factor each of the given polynomials completely. 

49. 12y 5 + 15y 4 - 108y 3 - 135y 2 53. 72z 6 + 108z 5 - 2z 4 - 3z 3 

50. 96 8 + 126 7 - 3246 6 - 4326 5 54. 216a; 7 + 324a; 6 - 6a; 5 - 9a; 4 

51. 9a; 6 ^ 5 + 6a; 5 ;z 6 - lUx 4 z 7 - 96x 3 z s 55. 144a 6 c 3 + 360a 5 c 4 - 4a 4 c 5 - 10a 3 c 6 

52. 12w 7 w 3 + 9u 6 w 4 - 432u 5 w 5 - 324w 4 w 6 56. 48a 8 c 4 + 32a 7 c 5 - 3a 6 c 6 - 2a 5 c 7 



In Exercises 57-60, use your calculator to help factor each of the given trinomials. Follow the procedure 
outline in Using the Calculator to Assist the ac-Method on page 462. 

57. 6a; 2 + 61a; + 120 59. 60a; 2 - 167a; + 72 

58. 16a; 2 - 62a; - 45 60. 28a; 2 + x - 144 



$»• 2** £*■ Answers ■*$ -as >*$ 

1. 4y 2 z 2 (lly + 6z)(lly - 6z) 13. 3a 4 (5a - 7) 2 

3. 3a; 5 z 5 (a; + ll)(a; - 11) 15 - 5a 3 6 3 (6a + 96) 2 

17. 26 3 (6+l) 2 
19. 2z 2 (z- l) 2 
21. 4a; 2 (9a; + 5) 2 
23. 36 2 c 5 (56-8c) 2 
11. 2w 3 w 3 (25u 2 + w 2 )(5u + w){5u - w) 25. 5a 3 (a - 6)(a + 7) 



5. 2u 5 {u + 9)(u-9) 

7. 3v 4 (v 2 + 25){v + 5)(v-5) 

9. 3a; 4 (x + 10) {x - 10) oq q^s/^ o„\2 



466 



27. 3y 4 (y-8)G/-5) 

29. 3z 2 (;z-5)0 + 9) 

31. 4a 4 (a + 9)(a + 7) 

33. 3z 2 (x + 7)(z + 4) 

35. 5z 5 (z-6)0-9) 

37. 26(26 -5)(6- 3) 

39. u 2 w 5 (u - 4w)(2w + 5w) 

41. 2x 2 y 5 {3x + by)(2x + by) 

43. 3x(4x - 5)(x + 2) 



CHAPTER 6. FACTORING 



45. 2u 4 (4u + 5)(w + 3) 
47. a 2 c 4 (4a-5c)(3o-5c) 
49. 3y 2 (y + 3)(y-3)(4 2 / + 5) 
51. 3x 3 z 5 (x + 4z)(x - 4z)(3x + 2z) 
53. z 3 (6z + l)(6z- l)(2z + 3) 
55. 2a 3 c 3 (6a + c)(6a - c)(2a + 5c) 
57. (2a! + 15)(3a; + 8) 
59. (15x-8)(4a;-9) 



6. 7. APPLICATIONS OF FACTORING 



467 



6.7 Applications of Factoring 

In this section we will solve applications whose solutions involve factoring. Let's 
begin. 



EXAMPLE 1. A rectangular canvas picture measures 12 inches by 16 inches. 
The canvas is mounted inside a frame of uniform width, increasing the total 
area covered by both canvas and frame to 396 square inches. Find the uniform 
width of the frame. 

Solution: We follow the Requirements for Word Problem Solutions. 

1. Set up a variable dictionary. A carefully labeled figure will help us main- 
tain our focus. We'll let x represent the uniform width of the frame. 



You Try It! 



A rectangular canvas picture 
measures 7 inches by 11 
inches. The canvas is 
mounted inside a frame of 
uniform width, increasing 
the total area covered by 
both canvas and frame to 
117 square inches. Find the 
uniform width of the frame. 



16 + 2a: 















X 








16 






X 


12 




12 


X 










16 










X 













Yl + 2x 



16 + 2a: 



2. Set up an equation. If the inner rectangular dimensions are 16 inches by 
12 inches, adding a frame of uniform width x makes the dimensions of 
the frame plus canvas 16 + 2x inches by 12 + 2x inches. The total area 
is found by multiplying these outer dimensions, A = (16 + 2x)(12 + 2x). 
If the total area is A = 396 square inches, then we have the following 
equation. 

(16 + 2x)(12 + 2a;) = 396 

3. Solve the equation. We start by expanding the right-hand side of the 
equation. 



(16 + 2a;)(12 + 2a;) = 396 
192 + 56z + Ax 2 = 396 



The resulting equation is nonlinear. Make one side zero. 



4oT + 56a; - 204 = 



468 



CHAPTER 6. FACTORING 



Answer: 1 inch 



We could factor out 4 on the left-hand side, but since there is a zero on 
the right-hand side of the equation, it's a bit easier to simply divide both 
sides by 4. Note how we divide each term on the left-hand side by the 
number 4. 

x 2 + 14a; -51 = 

We need an integer pair whose product is ac = —51 and whose sum is 
b = 14. The integer pair —3, 17 comes to mind. Since the coefficient of 
x 2 is one, we can simply "drop in place" our ordered pair. 

(z-3)(x + 17) = 

Thus, the solutions are x = 3 and x = —17. 

4. Answer the question. The uniform width of the frame cannot be —17 
inches, so we eliminate this solution from consideration. Hence, the uni- 
form width of the frame is 3 inches. 

5. Look back. If the uniform width of the frame is 3 inches, then the final 
dimensions will look like the following. 



22 





3 






16 






3 

t > 


12 


16 


12 


3 

< > 




3 



18 



22 



Thus, the combined area of the frame plus canvas is A = (18) (22), or 
A = 396 square inches, the area given in the problem statement. Our 
solution is correct. 



□ 



You Try It! 



A projectile's height (in feet) 
is given by the equation 
y= _i6t 2 + 144i + 576, 
where time t is measured in 
seconds. How much time 
passes before the projectile 
hits the ground? 



EXAMPLE 2. A projectile is fired at an angle into the air from atop a cliff 
overlooking the ocean. The projectile's distance (in feet) from the base of the 
cliff is give by the equation 

x = not, (6.1) 



6. 7. APPLICATIONS OF FACTORING 469 

and the projectile's height above sea level (in feet) is given by the equation 

y = _i6i 2 + 288i + 640, (6.2) 

where t is the amount of time (in seconds) that has passed since the projectile's 
release. 

a) How much time passes before the projectile splashes into the ocean? 

b) At that time, how far is the projectile from the base of the cliff? 

Solution: We follow the Requirements for Word Problem Solutions. 

1. Set up a variable dictionary. The variables are already set up for us. 

x = Distance from base of the cliff (in feet). 

y = Height above sea level (in feet). 

t = Time since projectile's release (in seconds). 

2. Set up an equation. The equations are already set up (see equation 6.1 
and equation 6.2). 

3. Solve the equation. When the projectile splashes into the ocean, its height 
above sea level at that moment is y = feet. Substitute for y in 
equation 6.2 and solve the resulting equation for t. 

y = _i6t 2 + 288i + 640 
= -I6i 2 + 288^ + 640 

We could factor out —16, but since the left-hand side of this equation is 
zero, it's a bit easier to divide both sides by —16. Note how we divide 
each term on the right-hand side by —16. 

= t 2 - 18* - 40 

We need a pair of integers so that their product is ac = —40 and their 
sum is —18. The integer pair 2, —20 comes to mind. Since the coefficient 
of t 2 is one, we can "drop in place" our integer pair. 

= (£ + 2)(£-20) 
Hence, the solutions are t = — 2 and t = 20. 

4. Answer the question. In answering question (a), the solution t = — 2 
seconds makes no sense. Thus, the projectile splashes into the ocean at 
t = 20 seconds. 



470 



CHAPTER 6. FACTORING 



Answer: 12 seconds 



In addressing question (b), to find the projectile's distance from the 
base of the cliff at this moment, substitute t = 20 in equation 6.1. 

x = 120* 
x = 120(20) 
x = 2400 

Hence, at the moment the projectile splashes into the ocean, it is 2,400 
feet from the base of the cliff. 

5. Look back. The best we can do here is check our solution * = 20 in 
equation 6.2. 

y = _i6* 2 + 288* + 640 
y = -16(20) 2 + 288(20) + 640 
y = -6400 + 5760 + 640 
y = 

Indeed, at * = 20, the projectile does splash into the ocean. 



□ 



You Try It! 



The product of two 
consecutive positive odd 
integers is 483. Find the 
integers. 



EXAMPLE 3. The product of two consecutive even integers is 728. Find 
the integers. 

Solution: We follow the Requirements for Word Problem Solutions. 

1. Set up a variable dictionary. Let k represent the first even integer. Then 
k + 2 represents the next consecutive even integer. 

2. Set up an equation. The product of the integers is 728. Hence, we have 
the following equation. 

k(k + 2) = 728 

3. Solve the equation. Expand the left-hand side of the equation. 

k 2 + 2k = 728 

The equation is nonlinear. Make one side zero. 

k 2 + 2k - 728 = 

See Using the Calculator to Assist the ac-Method on page 462. We need 
an integer pair whose product is ac = —728 and whose sum is b = 2. 
Enter -728/X in Yl, then set up the table (see Figure 6.31). 



6. 7. APPLICATIONS OF FACTORING 



471 



Plotl FlotE Plots 

WiB-728^X 

\Yi = 
We = 




TRBLE SETUP 

TblStart=l 

aTbl=l 

Indpnt: rang Rsk 
Depend: I3KSI Rsk 



X 


Vi 




21 


-30.33 




25 


"29.12 




26 


-SB 




2? 


"Efi.Sfi 




2B 


"Efi 




b 


"25.1 
-EH.E? 




X=3@ 



Figure 6.31: Load ac/X , or -728/X into Yl in the Y= menu. 



Use the up- and down-arrow keys to scroll. Note that 28, —26 is the 
desired pair. Because the coefficient of k 2 is one, we can "drop in place" 
the ordered pair. 

= (fc + 28)(fc-26) 

Hence, the solutions are k = —28 and k = 26. 

4. Answer the question. If k = —28, the next consecutive even integer is 
k + 2 = —26. Secondly, if k = 26, the next consecutive even integer is 

fc + 2 = 28. 

5. Lookback. Our first pair is —28 and —26. They have the required product 
of 728. Our second pair is 26 and 28. Their product is also 728. Both 
solutions check! 



Answer: 21 and 23 



□ 



EXAMPLE 4. A rectangle has perimeter 54 feet and area 180 square feet. 
Find the dimensions of the rectangle. 

Solution: We follow the Requirements for Word Problem Solutions. 



You Try It! 



A rectangle has perimeter 62 
feet and area 234 square feet. 
Find the dimensions of the 
rectangle. 



1. Set up a variable dictionary. A sketch will help us keep our focus. Let L 
represent the length of the rectangle and let W represent its width. 



W 



w 



472 



CHAPTER 6. FACTORING 



2. Set up an equation. The perimeter is 54 feet, so 2W+2L = 54, or dividing 
both sides by 2: 

W + L = 27 (6.3) 

The area is 180 square feet, so: 



LW = 180 



(6.4) 



3. Solve the equation. The system of equations (equations 6.3 and 6.4) can 
be solved using the substitution method. First, solve equation 6.3 for W: 

W = 27-L (6.5) 

Substitute equation 6.5 in equation 6.4, expand, then make one side zero. 

180 



L(27-L) 
27 X - X 2 



180 



= L 1 - 27L + 180 

The integer pair —12,-15 has product ac = 180 and sum b = —27. 
Moreover, the coefficient of X 2 is one, so we can "drop in place" our 
integer pair. 

= (X- 12) (X- 15) 

Hence, the solutions are X = 12 and X = 15. 
4. Answer the question. Two possibilities for the width. 

Substitute X = 12 in (6.5). Substitute L = 15 in (6.5). 



W = 27-L 

W = 27- 12 
W = 15 



W = 27-L 
W = 27-15 
W = 12 



Both answers give the same 15 by 12 foot rectangle, but we usually think 
of the term "length" as the longer side of the rectangle. So let's go with 
the length is L = 15 feet and the width is W = 12 feet. 

5. Look back. Let's add L = 15 feet and W = 12 feet to a diagram. 

15 ft 



12 ft 



12 ft 



15 ft 



6. 7. APPLICATIONS OF FACTORING 473 

If we add the dimensions around the rectangle, the perimeter is P = 

15 + 12 + 15 + 12, or P = 54 feet, the perimeter required in the problem 

statement. Next, if we multiply the dimensions, then A = (15)(12), or 

A = 180 square feet, the area required in the problem statement. Our Answer- length = 18 feet 

solution is correct! and width = 13 feet 



□ 



474 



CHAPTER 6. FACTORING 



f»- n- fa- 



Exercises 



■*1 ■*'» .*5 



1. A rectangular canvas picture measures 
14 inches by 36 inches. The canvas is 
mounted inside a frame of uniform width, 
increasing the total area covered by both 
canvas and frame to 720 square inches. 
Find the uniform width of the frame. 

2. A rectangular canvas picture measures 
10 inches by 32 inches. The canvas is 
mounted inside a frame of uniform width, 
increasing the total area covered by both 
canvas and frame to 504 square inches. 
Find the uniform width of the frame. 

3. A projectile is fired at an angle into the 
air from atop a cliff overlooking the ocean. 
The projectile's distance (in feet) from the 
base of the cliff is give by the equation 

x = 180£, 

and the projectile's height above sea level 
(in feet) is given by the equation 



y 



-16£ 2 + 352^ + 1664, 



where t is the amount of time (in seconds) 
that has passed since the projectile's re- 
lease. How much time passes before the 
projectile splashes into the ocean? At 
that time, how far is the projectile from 
the base of the cliff? 

4. A projectile is fired at an angle into the 
air from atop a cliff overlooking the ocean. 
The projectile's distance (in feet) from the 
base of the cliff is give by the equation 

x = UOt, 

and the projectile's height above sea level 
(in feet) is given by the equation 



where t is the amount of time (in seconds) 
that has passed since the projectile's re- 
lease. How much time passes before the 
projectile splashes into the ocean? At 
that time, how far is the projectile from 
the base of the cliff? 

5. The product of two consecutive even inte- 
gers is 624. Find the integers. 

6. The product of two consecutive even inte- 
gers is 528. Find the integers. 

7. The product of two consecutive positive 
integers is 552. Find the integers. 

8. The product of two consecutive positive 
integers is 756. Find the integers. 

9. The product of two consecutive odd inte- 
gers is 483. Find the integers. 

10. The product of two consecutive odd inte- 
gers is 783. Find the integers. 

11. A rectangle has perimeter 42 feet and area 
104 square feet. Find the dimensions of 
the rectangle. 

12. A rectangle has perimeter 32 feet and area 
55 square feet. Find the dimensions of the 
rectangle. 

13. The radius of the outer circle is one inch 
longer than twice the radius of the inner 
circle. 




V 



-16r + 288* + 1408, 



If the area of the shaded region is 407T 
square inches, what is the length of the 
inner radius? 



6. 7. APPLICATIONS OF FACTORING 



475 



14. The radius of the outer circle is two inches 
longer than three times the radius of the 
inner circle. 




15. 



If the area of the shaded region is 1807T 
square inches, what is the length of the 
inner radius? 

You have two positive numbers. The sec- 
ond number is three more than two times 
the first number. The difference of their 
squares is 144. Find both positive num- 
bers. 

You have two positive numbers. The sec- 
ond number is two more than three times 
the first number. The difference of their 
squares is 60. Find both positive num- 
bers. 

17. Two numbers differ by 5. The sum of their 
squares is 97. Find the two numbers. 



16. 



18. Two numbers differ by 6. The sum of their 
squares is 146. Find the two numbers. 

19. The length of a rectangle is three feet 
longer than six times its width. If the area 
of the rectangle is 165 square feet, what is 
the width of the rectangle? 

20. The length of a rectangle is three feet 
longer than nine times its width. If the 
area of the rectangle is 90 square feet, 
what is the width of the rectangle? 

21. The ratio of the width to the length of a 
given rectangle is 2 to 3, or 2/3. If the 
width and length are both increased by 4 
inches, the area of the resulting rectangle 
is 80 square inches. Find the width and 
length of the original rectangle. 

22. The ratio of the width to the length of a 
given rectangle is 3 to 4, or 3/4. If the 
width is increased by 3 inches and the 
length is increased by 6 inches, the area 
of the resulting rectangle is 126 square 
inches. Find the width and length of the 
original rectangle. 



j*- j*- j*- Answers •*$ **s •** 



1. 2 inches 

3. 26 seconds, 4,680 feet 
5. -26 and -24, and 24 and 26 
7. 23, 24 

9. -23 and -21, and 21 and 23 
11. 8 feet by 13 feet 



13. 3 inches 
15. 5 and 13 

17. 4 and 9, and -4 and -9 

19. 5 feet 

21. 4 inches by 6 inches 



Chapter 7 



Rational Expressions 



Our ability to communicate and record ideas is enhanced by concise mathe- 
matical language and notation. More importantly, our ability to conceive new 
ideas is broadened by notation that the mind can effectively use to organize 
and recognize patterns. Exponential notation is a simple example of such nota- 
tion that expands our thinking. The modern notation we use started seriously 
in 1636 and 1637 with James Humes and Rene Descartes, after centuries of 
mathematicians flirting with various approaches. Descartes may have been the 
first to use today's radical symbol, combining the check symbol with the bar 
over the top in 1637. Scientific notation, as we define it today, is a relatively 
new notation, first appearing in the mid-1900's. One may wonder where the 
next significant notational breakthrough will originate. Maybe it will be your 
contribution. 



477 



478 



CHAPTER 7. RATIONAL EXPRESSIONS 



7.1 Negative Exponents 

We begin with a seemingly silly but powerful definition on what it means to 
raise a number to a power of — 1 . 



Raising to a Power of —1. To raise an object to a power of —1, simply 
invert the object (turn it upside down). 



9 

A 



v 

6 



More formally, inverting a number is known as taking its reciprocal. 



You Try It! 



Simplify: ( - 



EXAMPLE 1. Simplify each of the following expressions: 

-1 / o\ -i 



a) 4- 



b) 



Solution: In each case, we simply invert the given number. 



a) A- 1 



1 



b) 



2\ X 3 



Solution: In each case, we simply invert the given number, 
a) 4- = i 



b) (-5)- 1 



1 



2\ 1 3 



d) 



3V 1 5 



Answer: — 

7 



□ 



7.1. NEGATIVE EXPONENTS 479 

You might be asking "Why does raising to the power of minus one invert 
the number?" To answer this question, recall the product of a number and its 
reciprocal is one. For example, 

4-i = l. (7.1) 

Next, consider what happens when we multiply 4 1 and 4 _1 . If we apply the 
usual law of exponents (assuming they work for both positive and negative 
exponents), we would add the exponents (1 + (—1) = 0). 

4 1 -4~ 1 =4° (7.2) 

However, because 4 1 = 4 and 4° = 1, this last equation is equivalent to: 

4-4 _1 = l (7.3) 

When you compare Equation 7.1 and 7.3, it is clear that 4 _1 and 1/4 are both 
reciprocals of the number 4. Because reciprocals are unique, 4 _1 = 1/4. 

In similar fashion, one can discover the meaning of a~ n . Start with the fact 
that multiplying reciprocals yields an answer of one. 

a n • 4 = 1 (7.4) 

If we multiply a n and a - ™, we add the exponents as follows. 

a n ■ a~ n = a 
Providing a ^ 0, then a = 1, so we can write 

a n -a- n = l (7.5) 

Comparing Equations 7.4 and 7.5, we note that both l/o n and a~" are recip- 
rocals of a n . Because every number has a unique reciprocal, a~ n and 1/a" are 
equal. 

Raising to a negative integer. Provided o/:0, 

a™ 



480 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



Simplify: 3 



Answer: 



EXAMPLE 2. Simplify each of the following expressions: 



>-3 



b) (-5)" 



(-4) 



-3 



Solution: In each example, we use the property a n = l/a n to simplify the 
given expression. 

a) 2- ' ' l l 



2 3 
1 



b) (-5)" 



(-5)2 
1 
25 



(-4)" 



(-4)3 

1 
~64 



In Raising to a Negative Integer on page 482, we'll address how you can per- 
form each of the above computations mentally. 

□ 



Laws of Exponents 

In the arguments demonstrating that 4 _1 = 1/4 and a~ n = 1/a™, we appealed 
to one of the laws of exponents learned in Chapter 5, Section 5. Fortunately, 
the laws of exponents work exactly the same whether the exponents are positive 
or negative integers. 



Laws of Exponents. If m and n are integers, then: 

1. a m a n = a m+n 

a m 

2. = a m - n 

a n 

3. ( a m ) n = a mn 

4. (ab) n = a n b n 



■- id' 



b n 



You Try It! 



Simplify: t s ■ V 



EXAMPLE 3. Simplify each of the following expressions: 

b) 2- 2 -2" 3 



a) y 5 y 7 



x~ 4 x 6 



Solution: In each case, we use the first law of exponents (a m a n = a m+n ). Be- 
cause we are multiplying like bases, we repeat the base and add the exponents. 



7.1. NEGATIVE EXPONENTS 



481 



= y~ 2 



b) 2- 2 • 2- 3 = 2- 2 +(- 3 > c) 



X -V = .*- 4 + 6 



Answer: £ 4 



D 



EXAMPLE 4. Simplify each of the following expressions: 



b) 



3 5 



Solution: In each case, we use the second law of exponents (a m /a n = a m ~ n ). 
Because we are dividing like bases, we repeat the base and subtract the expo- 
nents. Recall that subtraction means "add the opposite." 



.4-7 



X 



4+(-7) 



o-4 



-4-5 



-4+(-5) 



y-5 



-3-(-5) 



y -3+5 



You Try It! 



y 



Simplify: — - 

y-2 



Answer: y 



D 



EXAMPLE 5. Simplify each of the following expressions: 
a) (5- 2 ) 3 b) (a- 3 )- 4 c) (w 2 )" 7 

Solution: In each case, we are using the third law of exponents ((a m ) n = a mn ). 
Because we are raising a power to another power, we repeat the base and 
multiply the exponents. 



a) (5- 2 ) 3 =5(- 2 >( 3 > b) (a" 3 ) 



3\-4 



j(-3)(-4) 

,12 



C) («,') 



2\-7 



,(2)(-7) 



W 



-14 



You Try It! 



Simplify: (z ) 



5\-2 



Answer: 



a 



482 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



Simplify: ( - 



Raising to a Negative Integer 

We know what happens when you raise a number to — 1 , you invert the number 
or turn it upside down. But what happens when you raise a number to a 
negative integer other than negative one? 

As an example, consider the expression 3~ 2 . Using the third law of expo- 
nents ((a m )" = a" m ), we can write this expression in two equivalent forms. 

1. Note that 3 -2 is equivalent to (3 2 )" 1 . They are equivalent because the 
third law of exponents instructs us to multiply the exponents when raising 
a power to another power. Finally, note that to evaluate (S 2 )^ 1 , we first 



square, then invert the result. 



(3 2 ) 

9- 1 

1 
9 



Repeat base and multiply exponents. 



Simplify: 3 = 
Simplify: 9~ 



9. 



1/9. 



Note that 3~ 2 is also equivalent to (3 -1 ) 2 . They are equivalent because 
the third law of exponents instructs us to multiply the exponents when 
raising a power to another power. Finally, note that to evaluate (3 -1 ) 2 , 
we first invert, then square the result. 



(3 



-1\2 



Repeat base and multiply exponents. 
Simplify: 3" 1 = 1/3. 

Simplify: (1/3) 2 = 1/9. 



Using either technique, 3 -2 = 1/9. You can either square and invert, or you 
can invert and square. In each case, the 2 means "square" and the minus sign 
means "invert," and this example shows that it doesn't matter which you do 
first. 



EXAMPLE 6. Simplify each of the following expressions: 

-2 / r,\ -3 



a) 5~ 3 


b)(- 


_ 4 )-2 


Solution: 






a) We'll cube then invert 




5- 3 = 


(5 3 )- 1 






125 -1 
1 






125 



d) 



Repeat base, multiply exponents. 
Simplify: 5 3 = 125. 



Invert: 125" 



1/125. 



7.1. NEGATIVE EXPONENTS 483 

Note that the three means "cube" and the minus sign means "invert," so 
it is possible to do all of this work mentally: cube 5 to get 125, then invert 
to get 1/125. 

b) We'll square then invert. 

(— 4)~ = ((—4) )~ Repeat base, multiply exponents. 

= 16" 1 Simplify: (-4) 2 = 16. 

Invert: 16" 1 = 1/16. 



1 

16 



Note that the two means "square" and the minus sign means "invert," so 
it is possible to do all of this work mentally: square —4 to get 16, then 
invert to get 1/16. 



c) Again, we'll square then invert. 
3\ //3 X 



9 
25 

25 

~9~ 



Repeat base, multiply exponents. 

Simplify: (3/5) 2 = 9/25. 
Invert: (9/25)" 1 = 25/9. 



Note that the two means "square" and the minus sign means "invert," so 
it is possible to do all of this work mentally: square 3/5 to get 9/25, then 
invert to get 25/9. 



d) This time we'll cube then invert. 



2\ // 2 X 



3/ \ V 3. 

27 
27 



Repeat base, multiply exponents. 

Simplify: (-2/3) 3 = -8/27. 
Invert: (-8/27)" 1 = -27/8. 



Note that the three means "cube" and the minus sign means "invert," so 
it is possible to do all of this work mentally: cube —2/3 to get —8/27, then 
invert to get -27/8. 



64 

Answer: 

125 

□ 



484 



CHAPTER 7. RATIONAL EXPRESSIONS 



Applying the Laws of Exponents 

In this section we'll simplify a few more complicated expressions using the laws 
of exponents. 



You Try It! 



Simplify: 

(-5x s y- 2 )(-2x- 6 y- 1 ) 



EXAMPLE 7. Simplify: (2x- 2 y 3 ){-3x 5 y- 6 ) 

Solution: All the operators involved are multiplication, so the commutative 
and associative properties of multiplication allow us to change the order and 
grouping. We'll show this regrouping here, but this step can be done mentally. 

^arVX-SxV 6 ) = [(2)(-3)](a:- 2 x 5 )(y 3 y- 6 ) 

When multiplying, we repeat the base and add the exponents. 

= _6x- 2 +V+(- 6 ) 
= -6x 3 2T 3 

In the solution above, we've probably shown way too much work. It's far 
easier to perform all of these steps mentally, multiplying the 2 and the —3, 
then repeating bases and adding exponents, as in: 



Answer: lOx y 



2, ,-3 



„3„,-3 



(2x- 2 y i )(-3x b y" b ) = -Qx 6 y 



□ 



You Try It! 



Simplify: _ 

(4x z y°) 



fix v 
EXAMPLE 8. Simplify: „ _ 

9x 6 y 2 

Solution: The simplest approach is to first write the expression as a product. 

6x~ 2 y 5 6 x~ 2 y 5 
9x 3 y~ 2 9 x 3 y~ 2 

Reduce 6/9 to lowest terms. Because we are dividing like bases, we repeat the 
base and subtract the exponents. 



x -2- 3j/ 5-(-2) 



a ,-2+(-3) y 5+2 



X~V 



7.1. NEGATIVE EXPONENTS 



485 



In the solution above, we've probably shown way too much work. It's far easier 
to imagine writing the expression as a product, reducing 6/9, then repeating 
bases and subtracting exponents, as in: 



2„,5 



6x y 
9x 3 y~ 



f*-y 



Answer: — x 5 y 6 
2 y 



U 



You Try It! 



-2„,4\-3 



EXAMPLE 9. Simplify: (2arV) 

Solution: The fourth law of exponents {{ab) n = a n b n ) says that when you 
raise a product to a power, you must raise each factor to that power. So we 
begin by raising each factor to the minus three power. 



Simplify: (3x 4 j/ 



To raise two to the minus three, we must cube two and invert: 2 -3 = 1/8. 
Secondly, raising a power to a power requires that we repeat the base and 
multiply exponents. 



l a; (-2)(-3) y (4)(-3) 



;x V 



-12 



In the solution above, we've probably shown way too much work. It's far easier 
raise each factor to the minus three mentally: 2 -3 = 1/8, then multiply each 
exponent on the remaining factors by —3, as in 



(2z-V) 



-x 6 y- 12 



Answer: 



-x~ 8 y 6 



a 



Clearing Negative Exponents 

Often, we're asked to provide a final answer that is free of negative exponents. 
It is common to hear the instruction "no negative exponents in the final an- 
swer." Let's explore a couple of techniques that allow us to clear our answer 
of negative exponents. 



486 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



Simplify the expression 

x~ 2 

so that the resulting 
equivalent expression 
contains no negative 
exponents. 



EXAMPLE 10. Consider the expression: 

y-3 

Simplify so that the resulting equivalent expression contains no negative expo- 
nents. 

Solution: Raising y to the —3 means we have to cube and invert, so y~ 3 = 
1/y 3 . 

1 2 

X z X 



Answer: y 5 x 2 



y 



To divide x 2 by 1/y 3 , we invert and multiply. 



„2 . 



x 2 y 3 
1 I 



Alternate approach: An alternate approach takes advantage of the laws of 
exponents. We begin by multiplying numerator and denominator by y 3 . 



x 2 x 2 y 3 

y-3 y-3 y3 

x 2 y 3 



x 2 y 3 



In the last step, note how we used the fact that y = 1. 



a 



You Try It! 



Simplify the expression 
x~ 3 y 2 

so that the resulting 
equivalent expression 
contains no negative 
exponents. 



EXAMPLE 11. Consider the expression: 

2x 2 y~ 2 

z 3 

Simplify so that the resulting equivalent expression contains no negative expo- 
nents. 



7.1. NEGATIVE EXPONENTS 487 

Solution: Again, we can remove all the negative exponents by taking recip- 
rocals. In this case y~ 2 = 1/y 2 (square and invert). 

2 l 
2x 2 y~ 2 y 2 



2x 2 

yl 



To divide 2x jy by z , we invert and multiply. 



„2 



yl 


-z 3 


2x 2 


1 


V 2 


~ 


2x 2 




y 2 z 3 





Alternate approach: An alternate approach again takes advantage of the 
laws of exponents. We begin by multiplying numerator and denominator by 



y 2 - 



2x 2 y~ 


-2 


2x 2 y~ 2 


v 2 


z 3 




z 3 

2x 2 y° 
y 2 z 3 

2x 2 

y2 z 3 


y 2 



2 A 

V z 

In the last step, note how we used the fact that y = 1. Answer: — 

3x A 



D 



488 CHAPTER 7. RATIONAL EXPRESSIONS 



f> t* t* Exercises ■** -** ■** 



In Exercises 1-8, simplify the given expression. 



'•or 










4. 


nr 


/ 3\ _1 










5. 


(IS)" 1 


2 h) 










6. 


(-11)- 1 


( 8\ _1 










7. 


(16)- 1 


3 b) 










8. 


(7)- 1 


In Exercises 9-16, 


simplify 


the 


given 


expression. 






9. a~ 9 a 3 










13. 


2 9 . 2 -4 


10. :r~ 5 x~ 5 










14. 


2 2 -2- 7 


11. 6- 9 6 8 










15. 


9- 6 -9" 5 



12. v- 7 u- 2 16. 9 7 -9~ 5 



In Exercises 17-24, simplify the given expression. 

2 6 



17. 


2~8 


18. 


6 8 
6- 1 


19. 


z- 1 
z 9 


20. 


—4 


M> 3 



21. 


_9 

W 




w' 


22. 


r 5 
r -l 


23. 


7- 3 
7- 1 


24. 


6 -8 

6 6 



In Exercises 25-32, simplify the given expression. 

25. (r 1 ) 4 27. (6- 6 ) 7 

26. (a s y 7 28. (2- 7 )" 7 



7.1. NEGATIVE EXPONENTS 



489 



29. z" 



30. (c 6 Y 



31. (3- 2 ) : 

32. (8- 1 )* 



In Exercises 37-36, simplify the given expression. 



33. 4 3 

34. 5~ 2 

35. 2~ 4 

36. (-3)" 



38. 



39. 



37. 



40. 



In Exercises 41-56, simplify the given expression. 



41. (4u- 6 v~ 9 ) (5u 8 V 8 ) 



42. 


(6a- 9 c- 6 ) (-8a 8 c 5 ) 


43. 


(6x- 6 y- 5 ) (-4x 4 y- 


44. 


(5v- 3 w- s ) (8v- 9 w 5 


45. 


— 6x 7 z 9 
<±x~ 9 z~ 2 


46. 


2u~ 2 v 6 
6u 2 v 1 


47. 


-6a 9 c 6 
-4a" 5 c- 7 


48. 


— 4u~ 4 u; 4 



49. 


2v- 2 w 4 y'° 


50. 


3.s- 6 i 5 )" 4 


51. 


3x-V) 4 


52. 


-4 6- 8 c- 4 ) 3 


53. 


2x e z- 7 ) 5 


54. 


'-4v A w~ 9 ) 3 


55. 


2a- 4 C 8 )" 4 


56. 


ll^c- 1 ) -2 



In Exercises 57-76, clear all negative exponents from the given expression. 



57. 



58. 



59. 



x 5 y~ 


-2 


z 3 




4 - 


-9 


z 7 




Q — 

r s 


2 



p 



60. 



61. 



62. 



y 8 z 5 



y- 4 z 3 



490 CHAPTER 7. RATIONAL EXPRESSIONS 



63. — 1 —= 70 



64. 


a 7 
6- 8 c 6 


65. 


(7a;- 1 )(-7a;- 1 ) 


66. 


(3a- 8 )(-7a- 7 ) 


67. 


(8a- 8 )(7a- 7 ) 


68. 


(-7u 3 )(-8m- 6 ) 


fiq 


4a;" 9 



cS.r :i 



71 

72 



-6i 9 

6c 2 
-4c 7 

6v~ 9 



8v- 4 

73. (-3.s 9 )- 4 

74. (-3s 8 )- 4 

75. (2y 4 )- 5 

76. (2w 4 )- 5 



j*- j*- j*- Answers •*$ *** •** 



1. 7 27. 6 



7.1 
16 



13. 2 5 
15. 9- 11 



-42 



„ 9 29. z 81 



31. 3 6 
18 33. 



5.1 



64 

1 
35. — 

16 



9. a- 6 37. 32 

11. 6 * 39 - - 32 



41. 20u 2 «- 17 
43. -24a;- 2 y- 7 

45. --z 16 z n 
17. 2 14 2 

19.z-io 47.^V 3 

21. W - 16 49. l w io w -20 

32 

23 - 7 " 2 51. 81x- 4 y 28 

25. t- 4 53. 32x 30 z- 35 



7.1. NEGATIVE EXPONENTS 491 



55. 


-a 16 c' 32 




16 




X 5 


57. 


y 2 z 3 




„9 




r 


59. 






~&? 




x 3 y s 


61. 






z 5 




9 4 




U V 


63. 


w 7 




-49 


65. 


X 2 



67. 


56 


69. 


1 
2x 12 


71. 


3 

~2? 


73. 


1 


81s 36 


7c; 


1 



32y 



20 



492 



CHAPTER 7. RATIONAL EXPRESSIONS 



7.2 Scientific Notation 

We begin this section by examining powers of ten. 

10 1 = 10 

10 2 = 10-10= 100 

10 3 = 10- 10- 10 = 1,000 

10 4 = 10- 10- 10- 10= 10,000 

Note that the answer for 10 3 is a one followed by three zeros. The answer for 
10 4 is a one followed by four zeros. Do you see the pattern? 



Nonnegative powers of ten. In the expression 10™, the exponent matches 
the number of zeros in the answer. Hence, 10™ will be a 1 followed by n zeros. 



EXAMPLE 1. Simplify: 10 9 . 

Solution. 10 9 should be a 1 followed by 9 zeros. 

10 9 = 1,000,000,000 



□ 



You Try It! 



Simplify: 10 6 



Answer: 1,000,000 



Next, let's examine negative powers of ten. 

, 1 
10" 1 = — = 0.1 
10 



ur 



10" 



10" 



100 

1 

1000 

1 



10000 



0.01 

r = 0.001 

0.0001 



Note that the answer for 10 3 has three decimal places and the answer for 
10 -4 contains four decimal places. 



Negative powers of ten. In the expression 10 _n , the exponent matches 
the number of decimal places in the answer. Hence, 10 _ra will have n decimal 
places, the first n — 1 of which are zeros and the digit in the nth decimal place 
is a 1. 



7.2. SCIENTIFIC NOTATION 



493 



You Try It! 



EXAMPLE 2. Simplify: 1CT 7 . Simplify: 10~ 5 

Solution. 10 should have seven decimal places, the first six of which are 
zeros, and the digit in the seventh decimal place is a 1. 



itr 



0.0000001 



Answer: 0.00001 



□ 



Multiplying Decimal Numbers by Powers of Ten 

Let's multiply 1.234567 by 10 3 , or equivalently, by 1,000. 

1.234567 
xlOOO 



1234.567000 



The sum total of digits to the right of the decimal point in 1.234567 and 1000 
is 6. Therefore, we place the decimal point in the product so that there are six 
digits to the right of the decimal point. 

However, the trailing zeros may be removed without changing the value of 
the product. That is, 1.234567 times 1000 is 1234.567. Note that the decimal 
point in the product is three places further to the right than in the original 
factor. This observation leads to the following result. 

Multiplying by a nonnegative power of ten. Multiplying a decimal num- 
ber by 10™, where n = 0, 1, 2, 3, . . . , will move the decimal point n places to 
the right. 



EXAMPLE 3. Simplify: 325.6783 x 10 2 . 

Solution. Multiplying by 10 2 will move the decimal point two places to the 
right. Thus: 

325.6783 x 10 2 = 32,567.83 



You Try It! 



Simplify: 23.57889 x 10 3 



Answer: 23, 578.89 



□ 



EXAMPLE 4. Simplify: 1.25 x 10 5 



You Try It! 



Simplify: 2.35 x 10 4 



494 



CHAPTER 7. RATIONAL EXPRESSIONS 



Answer: 23, 500 



Solution. Multiplying by 10 5 will move the decimal point two places to the 
right. In this case, we need to add zeros at the end of the number to accomplish 
moving the decimal 5 places to the right. 



1.25 x 10 5 = 125,000 



□ 



Let's multiply 453.9 by 10 2 , or equivalently, by 0.01. 

453.9 
xO.01 



4.539 



The sum total of digits to the right of the decimal point in 453.9 and 0.01 is 3. 
Therefore, we place the decimal point in the product so that there are 3 digits 
to the right of the decimal point. That is, 453.9 x 10" 2 = 4.539. Note that 
the decimal point in the product is two places further to the left than in the 
original factor. This observation leads to the following result. 

Multiplying by a negative power of ten. Multiplying a decimal number 
by 10 _ ™, where n = 1, 2, 3, . . . , will move the decimal point n places to the 
left. 



You Try It! 



Simplify: 3, 854.2 x 10" 1 



Answer: 385.42 



EXAMPLE 5. Simplify: 14, 567.8 x 10" 3 . 

Solution. Multiplying by 10 -3 will move the decimal point three places to 
the left. Thus: 

14, 567.8 x 10~ 3 = 14.5678 

□ 



You Try It! 



Simplify: 2.2 x 10" 



EXAMPLE 6. Simplify: 4.3 x 10~ 4 . 

Solution. Multiplying by 10 -4 will move the decimal point four places to the 
left. In this case, we need to add some leading zeros at the beginning of the 
number to accomplish moving the decimal 4 places to the left. 



Answer: 0.022 



4.3 x 10" 



0.00043 



Note also the leading zero before the decimal point. Although .00043 is an 
equivalent number, the form 0.00043 is preferred in mathematics and science. 

□ 



7.2. SCIENTIFIC NOTATION 495 

Scientific Notation Form 

We start by defining the form of a number called scientific notation. 



Scientific notation 


A number having the form 












a x 


10 6 , 








where b is an integer 


and 1 < \a\ 


< 10, 


is said to be 


in 


scientific 


notation. 



The requirement 1 < \a\ < 10 says that the magnitude of a must be a least 1 
and less than 10. 

• The number 12.34 x 10 is not in scientific notation because |12.34| = 
12.34 is larger than 10. 

• The number —0.95 x 10 3 is not in scientific notation because | — 0.95| = 
0.95 is less than 1. 

• The number 7.58 x 10 -12 is in scientific notation because |7.58| = 7.58 is 
greater than or equal to 1 and less than 10. 

• The number —1.0 x 10 15 is in scientific notation because | — 1.0| = 1.0 is 
greater than or equal to 1 and less than 10. 

After contemplating these examples, it follows that a number in scientific no- 
tation should have exactly one of the digits 1, 2, 3, . . . , 9 before the decimal 
point. Exactly one, no more, no less. Thus, each of the following numbers is 
in scientific notation. 

4.7 xlO 8 , -3.764 xl0 _1 , 3.2x10°, and - 1.25 x 10 -22 

Placing a Number in Scientific Notation 

To place a number into scientific notation, we need to move the decimal point 
so that exactly one of the digits 1, 2, 3, . . . , 9 remains to the left of the 
decimal point, then multiply by the appropriate power of 10 so that the result 
is equivalent to the original number. 



You Try It! 



EXAMPLE 7. Place the number 1,234 in scientific notation. Place the number 54,321 

Solution. Move the decimal point three places to the left so that it is posi- 
tioned just after the 1. To make this new number equal to 1,234, multiply by 
10 3 . Thus: 

1,234= 1.234 x 10 3 



496 



CHAPTER 7. RATIONAL EXPRESSIONS 



Answer: 5.4321 x 10 4 



Check: Multiplying by 10 3 moves the decimal three places to the right, so: 

1.234 x 10 3 = 1,234 
This is the original number, so our scientific notation form is correct. 



□ 



You Try It! 



Place the number 0.0175 EXAMPLE 8. Place the number 0.000025 in scientific notation. 

Solution. Move the decimal point five places to the right so that it is posi- 
tioned just after the 2. To make this new number equal to 0.000025, multiply 
by 10" 5 . Thus: 

0.000025 = 2.5 x 10~ 5 

Check: Multiplying by 10 -5 moves the decimal five places to the left, so: 

2.5 x 10~ 5 = 0.000025 
Answer: 1.75 x 10 -2 This is the original number, so our scientific notation form is correct. 

□ 



You Try It! 



Place the number 
756.98 x 10" 5 in scientific 
notation. 



EXAMPLE 9. Place the number 34.5 x 10 n in scientific notation. 

Solution. First, move the decimal point one place to the left so that it is 
positioned just after the three. To make this new form equal to 34.5, multiply 
by 10 1 . 

34.5 x 10~ n = 3.45 x 10 1 x 10~ n 
Now, repeat the base 10 and add the exponents. 



3.45 x 10 



-10 



Answer: 7.5698 x 10" 



□ 



You Try It! 



Place the number 
0.00824 x 10 8 in scientific 
notation. 



EXAMPLE 10. Place the number 0.00093 x 10 12 in scientific notation. 

Solution. First, move the decimal point four places to the right so that it 
is positioned just after the nine. To make this new form equal to 0.00093, 
multiply by 10~ 4 . 



0.00093 x 10 



12 



9.3 x 10" 4 x 10 12 



7.2. SCIENTIFIC NOTATION 



497 



Now, repeat the base 10 and add the exponents. 

= 9.3 x 10 8 



Answer: 8.24 x 10 5 



Scientific Notation and the Graphing Calculator 

The TI-84 graphing calculator has a special button for entering numbers in 
scientific notation. Locate the "comma" key just about the number 7 key 
on the calculator's keyboard (see Figure 7.1). Just above the "comma" key, 
printed on the calculator's case is the symbol EE. It's in the same color as the 
2nd key, so you'll have to use the 2nd key to access this symbol. 




□ 



Figure 7.1: Locate the EE label above the "comma" key on the keyboard. 



We know that 2.3 x 10 2 = 230. Let's see if the calculator gives the same 
interpretation. 

1. Enter 2.3. 

2. Press the 2nd key, then the comma key. This will put E on the calculator 
view screen. 

3. Enter a 2. 

4. Press ENTER. 

The result of these steps is shown in the first image in Figure 7.2. Note that 
the calculator interprets 2.3E2 as 2.3 x 10 2 and gives the correct answer, 230. 
You can continue entering numbers in scientific notation (see the middle image 
in Figure 7.2). However, at some point the numbers become too large and 
the calculator responds by outputting the numbers in scientific notaiton. You 
can also force your calculator to display numbers in scientific notation in all 
situations, by first pressing the MODE key, then selecting SCI on the first 



498 



CHAPTER 7. RATIONAL EXPRESSIONS 



2.3e2 



230 



2. 


3e5 


230000 


2. 


3e10 


2.3e10 




IriDKHHL HB EDG 

(i 1 i ]: H E b ? B 5 
DEGREE 
PAR PDL SE9 
DDT 
SIHUL 
a+bi- rt A $i< 
HDRIZ G-T 
inEXTl 



Figure 7.2: Entering numbers in scientific notation. 



You Try It! 



Use the graphing calculator 
to simplify: 

(3.42 x 10 6 )(5.86 x 1CT 9 ) 



line and pressing the ENTER key (see the third image in Figure 7.2). You can 
return your calculator to "normal" mode by selecting NORMAL and pressing 
the ENTER key. 



EXAMPLE 11. Use the graphing calculator to simplify: 

(2.35 x 1CT 12 )(3.25 x 1(T 4 ) 

Solution. First, note that we can approximate (2.35 x 10 -12 )(3.25 x 10 -4 ) by 
taking the product of 2 and 3 and adding the powers of ten. 

(2.35 x 1(T 12 )(3.25 x 10" 4 ) 

w (2 X 1(T 12 )(3 x 1(T 4 ) Approximate: 2.35 « 2 and 3.25 « 3. 
w 6 x 10~ 16 2 • 3 = 6 and 10" 12 ■ 10" 4 = 10" 16 . 

The graphing calculator will provide an accurate answer. Enter 2.35E-12, 
press the "times" button, then enter 3.25E-4 and press the ENTER button. 
Be sure to use the "negate" button and not the "subtract" button to produce 
the minus sign. The result is shown in Figure 7.3. 



2.35e-12*3.25e-4 
7.6375E-16 



Figure 7.3: Computing (2.35 x 1(T 12 )(3.25 x 10" 



Answer: 2.00412 x 10" 



Thus, (2.35 x 10~ 12 )(3.25 x 10~ 4 ) 
close to our estimate of 6 x 10~ 16 . 



7.6375 x 10" lb . Note that this is fairly 



□ 



7.2. SCIENTIFIC NOTATION 



499 



Reporting your answer on your homework. After computing the answer 
to Example 11 on your calculator, write the following on your homework: 

(2.35 x 1CT 12 )(3.25 x 10" 4 ) = 7.6375 x 1CT 16 

Do not write 7.6375E-16. 



You Try It! 



EXAMPLE 12. Use the graphing calculator to simplify: 

6.1 x 1(T 3 



(2.7 x 10 4 )(1.1 x 10 8 ) 



Use the graphing calculator 
to simplify: 

2.6 x 10 4 
(7.1 x 10" 2 )(6.3x 10 7 ) 



Solution. Again, it is not difficult to produce an approximate answer. 
6.1 x 10~ 3 



(2.7 x 10 4 )(1.1 x 10 8 ) 

6 x 10" 3 
~ (3 x 10 4 )(1 x 10 8 ) 
Jx 10" 3 
" 3 x 10 12 
- 6 10~ 3 
~ 3 'TniT 



.1«6, 2.7 « 3, and 1.1 w 1. 



10 



2 x 10 



-15 



3 • 1 = 3 and 10 4 • 10 8 = 10 



ac a c 

bd = b'd' 

6 10~ 3 

- = 2 and — — = 10" 

3 10 12 



12 



-15 



Let's get a precise answer with our calculator. Enter the numerator as 6.1E- 
3, then press the "division" button. Remember that we must surround the 
denominator with parentheses. So press the open parentheses key, then enter 
2.7E4. Press the "times" key, then enter 1.1E8. Press the close parentheses 
key and press the ENTER button. The result is shown in Figure 7.4. 



6.1e-3/<2.7e4*1. 
1eS> 

2.053S7205E-15 



Figure 7.4: Computing 6.1 x 10" 3 /(2.7 x 10 4 x 1.1 x 10 



Thus, 6.1 x 10~ 3 /(2.7 x 10 4 x 1.1 x 10 8 ) = 2.05387205 x 10" 15 . Note that 
this is fairly close to our estimate of 2 x 10 -15 . 



Answer: 5.8126537 x 10" 



□ 



500 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



The mass of the 
International Space Station 
is 450,000 kg, and its average 
distance to the center of the 
earth is 387,000 m. Find the 
force of attraction between 
the earth and the station 
(in newtons (N)). 



EXAMPLE 13. Isaac Newton's universal law of gravitation is defined by 

the formula 

GmM 

where F is the force of attraction between two objects having mass m and 
M, r is the distance between the two objects, and G is Newton's gravitational 
constant defined by: 

G = 6.67428 x 10" 11 N(m/kg) 2 

Given that the mass of the moon is 7.3477 x 10 22 kilograms (kg), the mass of 
the earth is 5.9736 x 10 24 kilograms (kg), and the average distance between the 
moon and the earth is 3.84403 x 10 8 meters (m), find the force of attraction 
between the earth and the moon (in newtons (N)). 

Solution. Plug the given numbers into Newton's universal law of gravitation. 

GmM 
F = - — 

F 



(6.673 x 10~ n )(7.3477 x 10 22 )(5.9736 x 10 24 ) 



(3.84403 x 10 8 ) 2 
Enter the expression into your calculator (see Figure 7.5) as: 

(6.673E-11*7.3477E22*5.9736E24)/(3.84403E8)A2. 



Answer: 



1.20 x 10 9 N 



C6.673E -11*7. 347 
7e22*5.9736e24V 
(3.84403 £8^2 
1.9821 43728 e2S 



Figure 7.5: Computing force of attraction between earth and the moon. 

Hence, the force of attraction between the earth and the moon is approxi- 
mately 1.98 x 10 20 newtons (N). 

□ 



You Try It! 



The star Sirius is 8.58 
light-years from the earth. 
How many miles from the 
earth is Sirius? 



EXAMPLE 14. The closest star to the earth is Alpha Centauri, 4.37 light- 
years from the earth. A light-year is the distance that light will travel in 



7.2. SCIENTIFIC NOTATION 



501 



one-year's time. The speed of light is 186,000 miles per second. How many 
miles from the earth is Alpha Centauri? 

Solution: Because the speed of light is measured in miles per second, let's 
first compute the number of seconds in 4.37 years. Because there are 365 days 
in a year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, 
we can write: 

day hr min s 

4.37 yr = 4.37yr x 365 — x 24 - — x 60 - — x 60 — — 
yr day hr min 

day' faf rain s 

= 4.37 yr x 365 — - x 24 -— x 60 ±=-r x 60 — ^ 
yf pay .far jam 

Note how the units cancel, indicating that the final answer is in seconds. With 
our calculator mode set to scientific notation (see the image on the right in 
Figure 7.2), we multiply the numbers to get the result shown in Figure 7.6. 
Rounding, the number of seconds in 4.37 years is approximately 1.38 x 10 8 
seconds. 

Next, we compute the distance the light travels in 4.37 years. Using the 
fact that the distance traveled equals the speed times the time traveled, we 
have: 

Distance = Speed x Time 

= 1.86 x 10 5 — • 1.38 x 10 8 s 
s 

= 1.86 x 10 5 ^ • 1.38 x lO 8 * 1 
r 
Note how the units cancel, indicating that our answer is in miles. Again, 
with our calculator set in scientific notation mode, we compute the product of 
1.86 x 10 5 and 1.38 x 10 8 . The result is shown in the image on the right in 
Figure 7.6. 



4.37*365*24*60*6 


1. 3781232 e8 



1.86e5*1.33e8 

2.5668e13 



Figure 7.6: Computing the distance to Alpha Centauri in miles. 



Thus, the star Alpha Centauri is approximately 2.5668 x 10 13 miles from 
the earth, or 

2.5668 x 10 13 miles w 25, 668, 000, 000, 000 miles, 

pronounced "twenty-five quadrillion, six hundred sixty-eight trillion miles." 



Answer: 

« 5.2425 x 10 13 miles 



□ 



502 CHAPTER 7. RATIONAL EXPRESSIONS 

**> t* t* Exercises ■** ■*$ •*$ 

In Exercises 1-8, write each of the following in decimal format. 

1. 10" 4 5. 10 8 

2. 1(T 13 6. 10 14 

3. 1CT 8 7. 10 7 

4. 10" 9 8. 10 9 



In Exercises 9-16, write each of the following in decimal format. 

9. 6506399.9 x 10~ 4 13. 440906.28 x 10" 4 

10. 19548.4 x 10~ 2 14. 9147437.4 x 10~ 4 

11. 3959.430928 x 10 2 15. 849.855115 x 10 4 

12. 976.841866 x 10 4 16. 492.4414 x 10 3 



In Exercises 17-24, convert each of the given numbers into scientific notation. 

17. 390000 21. 0.81 

18. 0.0004902 22. 83400 

19. 0.202 23. 0.0007264 

20. 3231 24. 0.00395 

In Exercises 25-32, convert each of the given expressions into scientific notation. 

25. 0.04264 x 10~ 4 29. 30.04 x 10 5 

26. 0.0019 x 10" 1 30. 76000 x 10" 1 

27. 130000 x 10 3 31. 0.011 x 10 1 

28. 738 x lO" 1 32. 496000 x 10~ 3 



7.2. SCIENTIFIC NOTATION 



503 



In Exercises 33-38, each of the following numbers are examples of numbers reported on the graphing 
calculator in scientific notation. Express each in plain decimal notation. 



33. 1.134E-1 

34. 1.370E-4 

35. 1.556E-2 



36. 1.802E4 

37. 1.748E-4 

38. 1.402E0 



In Exercises 39-42, first, use the technique of Example 11 to approximate the given product without 
the use of a calculator. Next, use the MODE button to set you calculator in SCI and FLOAT mode, 
then enter the given product using scientific notation. When reporting your answer, report all digits 
shown in your calculator view screen. 



39. (2.5 x 10 _1 )(1.6 x 1(T 7 ) 

40. (2.91 x 10" 1 )(2.81 x 1(T 4 ) 



41. (1.4 x 10 7 )(1.8 x 1CT 4 ) 

42. (7.48 x 10 7 )(1.19 x 10 6 ) 



In Exercises 43-46, first, use the technique of Example 12 to approximate the given quotient without 
the use of a calculator. Next, push the MODE button, then highlight SCI mode and press ENTER. 
Move your cursor to the same row containing the FLOAT command, then highlight the number 2 
and press ENTER. This will round your answers to two decimal places. Press 2nd MODE to quit the 
MODE menu. With these settings, enter the given expression using scientific notation. When entering 
your answer, report all digits shown in the viewing window. 



3.2 x 1CT 5 
43. = 

2.5 x 10- 7 

6.47 x 1CT 5 

44 

1.79 x 10 8 



45 



46 



5.9 x 10 3 
2.3 x 10 5 

8.81 x 10" 9 
3.06 x lO" 1 



47. Overall the combined weight of biolog- 
ical material - animals, plants, insects, 
crops, bacteria, and so on - has been es- 
timated to be at about 75 billion tons 
or 6.8 x 10 13 kg (http: //en.wikipedia. 
org/wiki/Nature). If the Earth has mass 
of 5.9736 x 10 24 kg, what is the percent 
of the Earth's mass that is made up of 
biomass? 



48. The Guinness World Record for the 
longest handmade noodle was set on 
March 20, 2011. The 1,704-meter- long 
stretch of noodle was displayed during 
a noodle-making activity at a square 
in Southwest China's Yunnan province. 
Meigan estimates that the average width 
of the noodle (it's diameter) to be the 
same as her index finger or 1.5 cm. Using 
the volume formula for a cylinder (V = 
irr 2 h) estimate the volume of the noodle 
in cubic centimeters. 



504 



CHAPTER 7. RATIONAL EXPRESSIONS 



49. Assume there are 1.43 x 10 6 miles of paved 
road in the United States. If you could 
travel at an average of 65 miles per hour 
nonstop, how many days would it take you 
to travel over all of the paved roads in the 
USA? How many years? 



50. The population of the USA in mid-2011 
was estimated to be 3.12 x 10 8 people and 
the world population at that time to be 
about 7.012 x 10 9 people. What percent- 
age of the world population live in the 
USA? 



j*- j*- j*- Answers •*$-*$■*$ 



1. 0.0001 
3. 0.00000001 
5. 100,000,000 
7. 10,000,000 
9. 650.63999 
11. 395943.0928 
13. 44.090628 
15. 8498551.15 
17. 3.9 x 10 5 
19. 2.02 x lO" 1 
21. 8.1 x lO" 1 
23. 7.264 x 10~ 4 
25. 4.264 x 10~ 6 



27. 1.3 x 10 8 
29. 3.004 x 10 6 
31. 1.1 x 10" 1 
33. 0.1134 
35. 0.01556 
37. 0.0001748 
39. 4 x 10~ 8 
41. 2.52 x 10 3 
43. 1.28 x 10 2 
45. 2.57 x 10~ 2 
47. 1.14 x 10" 11 
49. 916.7 days, 2.5 yr 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 



505 



7.3 Simplifying Rational Expressions 

Any time you divide a polynomial by a second polynomial, you form what is 
known as a rational expression. 



Rational expression. The expression 

P{x) 
q(x) } 

where p(x) and q(x) are polynomials, is called a rational expression. 



Readers are strongly 
encouraged to review the 
material on fractions 
presented in Section 3 of 
Chapter 1. You will find 
that material quite helpful 
for this section. 



For example, each of the following is a rational expression. 

■2 , s x + 3 , 2x 

c) v 



3.7.' 



b) 



2a; -4 



In example a) , the rational expression is composed of a binomial over a mono- 
mial. Example b) is constructed by dividing a binomial by a trinomial. Ex- 
ample c) is composed of a monomial over a monomial, the type of rational 
expression that will gain the most attention in this section. 

Multiplying and Dividing Rational Expressions 

We will concentrate on rational expressions with monomial numerators and 
denominators. Recall that to form the product of two rational numbers, we 
simply multiply numerators and denominators. The same technique is used to 
multiply any two rational expressions. 

Multiplying rational expressions. Given a/b and c/d, their product is 

defined as: 

a c ac 

~b"d = bd 



Remember, you need only multiply numerators and denominators. For exam- 
ple: 



x 2 
3'y 



2.7' 

3? 



2a 5a 
36 1 ' g^3 



10a 2 
27&5' 



and 



x 

2y 



3x 

4y2 



3x 2 
8^ 



Of course, as the next example shows, sometimes you also need to reduce 
your answer to lowest terms. 



EXAMPLE 1. Simplify: 



You Try It! 



Simplify: 



9 x 
x 2 6 



506 



CHAPTER 7. RATIONAL EXPRESSIONS 



Answer: — 
2x 



Solution: Multiply numerators and denominators. 

2 x 2 _ 2x b 
x 4 4x 3 

Now, there several different ways you can reduce this answer to lowest terms, 
two of which are shown below. 

You can factor numerator and de- Or you can write the answer as a 
nominator, then cancel common product, repeat the base and sub- 



factors. 



tract exponents. 



2a; 5 
4a; 3 



2 ■ x ■ x ■ x ■ x ■ x 
2 ■ 2 ■ x ■ x ■ x 

% ■ ft ■ ft ■ ft ■ X ■ X 

_x^ 
2 

As dividing by 2 is the same as multiplying by 1/2, these answers are equivalent. 
Also, note that the right-hand method is more efficient. 

□ 



2a; 5 




2 


X 5 


4x 3 


"" 


4 


X 3 




= 


1 

2 


x 5 - 




= 


1 2 

r 



Recall that when dividing fractions, we invert the second fraction and mul- 
tiply. 



Dividing 


rational 


expressions. 


Given 


a/b and 


c/d, their quotient 


is defined 


as: 




a 
b 


c 

~ d ~ 


a d 

b'~c 
ad 

~b~c 







You Try It! 



Simplify: 



3y ^ r 

a; 3 4a; 



2 4 

EXAMPLE 2. Simplify: — -=- At 




V 2y^ 




Solution: Invert, then multiply. 




1 A 

X _ X 


x 2 2y 2 


y ' 2y 2 


y x A 




2x 2 y 2 




x 4 y 



Now, there several different ways you can reduce this answer to lowest terms, 
two of which are shown below. 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 



507 



You can factor numerator and de- 
nominator, then cancel common 
factors. 

2x 2 y 2 
x A y 



Or you can write the answer as a 
product, repeat the base and sub- 
tract exponents. 



2 ■ x ■ x ■ y 


!J 


x ■ x ■ X ■ X 


y 
y 



2„,2 



1x z y 
x 4 y 



x 4 y 1 



ft ■ ft ■ X ■ X ■ 

2y 



2x- 2 y l 

2y 



In the last step, x~ 2 is the same as 
1/x 2 , then we multiply numerators 
and denominators. 



Note that the right-hand method is more efficient. 



Answer: 



12 
x 2 y 



Adding and Subtracting Rational Expressions 

First, recall the rules for adding or subtracting fractions that have a "common" 
denominator. 



Adding rational expressions. Given a/c and b/c, their sum is defined as: 

a b a + b 
c c c 

That is, add the numerators and place the result over the common denominator. 



The following examples each share a common denominator. We add the nu- 
merators, then place the result over the common denominator. 



□ 



5 1 _ 6 2 3_5 

7 7 7' x x x' 



and 



x 3y x + 3y 

y y y 



3a: 2v 
EXAMPLE 3. Simplify: — + — 

xy xy 

Solution: Add the numerators, placing the result over the common denomi- 
nator. 

3a; 2y 3a; + 2y 



xy xy 



xy 



You Try It! 



Simplify: — 1 r— 

x^y x z y 



Answer: 



4a; + by 2 



x 2 y 



□ 



508 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



Simplify: 



8.r 



2x 



3yz 2 3yz 2 



Subtracting rational expressions. Given a/c and b/c, their difference is 

defined as: 

a b a — b 



c c c 

That is, subtract the numerators and place the result over the common denom- 
inator. 

The following examples each share a common denominator. We subtract the 
numerators, then place the result over the common denominator. 



5a 
T 



3o 

T 



2a 



and 



3;r 
xy 



xy 



3x — by 
xy 



As the next example shows, sometimes you may have to reduce your answer 
to lowest terms. 



EXAMPLE 4. Simplify: — - — 

Solution: Subtract the numerators, placing the result over the common 
denominator. 

hxy 3xy bxy — 3xy 

~z 2z ~ 2~z 

_ 2xy 

2z 
To reduce to lowest terms, divide both numerator and denominator by 2. 



Answer: 



2x 

yz 2 



xy_ 
z 



□ 



The Least Common Denominator 

When adding or subtracting, if the rational expressions do not share a com- 
mon denominator, you must first make equivalent fractions with a common 
denominator. 



Least common denominator. If the fractions a/b and c/d do not share 
a common denominator, then the least common denominator for b and d is 
defined as the smallest number (or expression) divisible by both b and d. In 
symbols, LCD(6, d) represents the least common denominator of b and d. 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 



509 



You Try It! 



EXAMPLE 5. 



Simplify: 1 

1 J 6 9 



Simplify: — 



5:)' 



Solution: The smallest number divisible by both 6 and 9 is 18; i.e., LCD(6, 9) = 
18. We must first make equivalent fractions with a common denominator of 
18. 



x 2x 
6 + Y 


a; 3 2a; 

~ 6 ' 3 + ~9~ 

3a; 4a; 

~ 18 + 18 


2 
2 



We can now add the numerators and put the result over the common denomi- 
nator. 

- 7 JL 

~ 18 



Answer: 



29.x 
~2A 



□ 



You Try It! 



EXAMPLE 6. 



Simplify: JL - JL 



Solution: The smallest expression divisible by both 8a; and 12a; is 24a;; 
i.e., LCD(8a;, 12a;) = 24a;. We must first make equivalent fractions with a 
common denominator of 24a;, then place the difference of the numerators over 
the common denominator. 



Simplify: 

8y lOy 



y 

8x ' 


y _ y 3 y 

" 12a; ~ 8a; ' 3 12a; 
_ 3y 2y 

24a; 24a; 

y 

24a; 


2 
2 



Answer: 



40y 



D 



In Example 5, it was not difficult to imagine the smallest number divisible 
by both 6 and 9. A similar statement might apply to Example 6. This is not 
the case in all situations. 



EXAMPLE 7. Simplify: 



21L 

72 



y 

108 



You Try It! 



Simplify: 



7x 
36 



3.r 

40 



510 



CHAPTER 7. RATIONAL EXPRESSIONS 



Solution: In this example, it is not easy to conjure up the smallest number 
divisible by both 72 and 108. As we shall see, prime factorization will come to 
the rescue. 



72 



108 



27 



Thus, 72 = 2 3 ■ 3 2 and 108 = 2 2 • 3 3 . 

Procedure for finding the least common denominator (LCD). To find 
the least common denominator for two or more fractions, proceed as follows: 

1. Prime factor each denominator, putting your answers in exponential form. 

2. To determine the LCD, write down each factor that appears in your prime 
factorizations to the highest power that it appears. 



Following the procedure above, we list the prime factorization of each denom- 
inator in exponential form. The highest power of 2 that appears is 2 3 . The 
highest power of 3 that appears is 3 3 . 



Answer: 



43a; 
360 



72 

108 

LCD 



2 3 -3 2 
2 2 -3 3 
2 3 -3 3 



Therefore, the LCD is 2 3 • 3 3 



hy 
72 " 


y _5y 3 y 
' 108 ~~ 72 ' 3 108 
_ 15y 2y 
216 216 
_ 13y 
216 


2 

' 2 



Prime factor 72. 

Prime factor 108. 

Highest power of 2 is 2 . 
Highest power of 3 is 3 . 

27 or 216. Hence: 



Make equivalent fractions. 

Simplify. 

Subtract numerators. 



□ 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 



511 



EXAMPLE 8. Simplify: 



7 



11 



15xy 2 20a; 2 

Solution: Prime factor each denominator, placing the results in exponential 
form. 

15xy = 3 • 5 ■ x ■ y 
20a; 2 = 2 2 ■ 5 • x 2 

To find the LCD, list each factor that appears to the highest power that it 
appears. 



You Try It! 



Simplify: 



11 



7x 



I8x 2 y 30xy 



2 „,2 



LCD = 2 Z ■ 3 • 5 • x z ■ y 



Simplify. 



LCD = 60a;V 

After making equivalent fractions, place the difference of the numerators over 
this common denominator. 



7 



11 



7 Ax 11 3y 2 
I5xy 2 20x 2 lbxy 2 Ax 20x 2 3y 2 



28a; 



33y 2 



60x 2 y 2 60x 2 y 2 
28a; - 33y 2 
60x 2 y 2 



Answer: 



55 + 21a; J 
90x 2 y 



D 



Dividing a Polynomial by a Monomial 

We know that multiplication is distributive with respect to addition; that is, 
a(b + c) = ab + ac. We use this property to perform multiplications such as: 



x 2 (2a; 2 -3a;- 



2a; 4 - 3x d 



8x z 



However, it is also true that division is distributive with respect to addition. 



512 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



Divide 9x 3 - 
by 3a; 2 . 



8a; — 6a; 



For example, note that 



4 + 6 _ 4 6 

2 ~ 2 + 2' 



This form of the distributive property can be used to divide a polynomial by 
a monomial. 



EXAMPLE 9. Divide x 2 - 2x - 3 by x 2 . 

Solution: We use the distributive property, dividing each term by x 2 . 



Answer: 3a; H 

3 x 



Now we reduce each term of the last result to lowest terms, canceling common 
factors. 

2 3 



□ 



You Try It! 



Divide -Ax 2 + 6x - 9 
by 2a; 4 . 



2 3 9 

Answer: -\ 



EXAMPLE 10. Divide 2x 3 - 3a; + 12 by 6a; 3 . 

Solution: We use the distributive property, dividing each term by 6a; 3 . 

2a; 3 - 3a; + 12 _ 2x 3 3x 12 
6a; 3 6a; 3 6x 3 6a; 3 

Now we reduce each term of the last result to lowest terms, canceling common 
factors. 



1 



1 

2^2 



□ 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 513 



**>**■**> Exercises ■** * «•* 



In Exercises 1-8, simplify each of the given experssions. 





12 


s 5 


1. 


T 2 ' 


9 


2. 


6 

T 4 


a: 2 
' 10 




12 


v 4 


3. 


y 3 


10 


4. 


10 
¥' 


i 5 
12 





s 5 


9s 2 


b. 


t 4 ' 


" t 2 




s 2 


6s 4 


6. 


£ 2 ' 


' t 4 




6 4 


96 2 


7. 


c 4 ' 


"T 2 " 




6 5 


86 2 


8. 


c 4 ' 


' c 2 



In Exercises 9-14, simplify each of the given expressions. 



9. 


10s 19s 

~T8~ + Ts" 


10. 


Uy lOy 


11. 


5 17 
9c ~ 9c 



12. 


19 
147 ~ 


17 
147 


13. 


8a; 
\byz 


16a; 
15j/z 


1/1 


17a 


9a 



206c 206c 



In Exercises 15-20, simplify each of the given expressions. 



15. 


9z 5z 
10 + T 


16. 


7u llu 


17. 


3 4 
10w 5v 



9 
18. 


7 


lOv 


2v 


8r 
19. 

5st 


9r 
lOst 


7a; 


3.r 


6y^ 


2 W z 



In Exercises 21-32, simplify each of the given expressions. 

11 5 5 17 

21. k + —^ 23. 



18rs 2 24r 2 s ' 24rs 2 36r 2 s 

5 13 13 19 

22. 24. 

Yluw 1 hAu 2 w 54ww 2 2Av 2 w 



514 CHAPTER 7. RATIONAL EXPRESSIONS 

7 11 

25. - H - 29. 

36y 3 48z 3 

™ 19 5 

26 36^ + 48? 3 °- 

27.^ + ^- 31. 

48w 3 36u> 3 

7 17 
28 1 32 

72r 3 48s 3 50rs 40st 



In Exercises 33-48, use the distributive property to divide each term in the numerator by the term in 
the denominator. 



11 


9 


50xy 


AOyz 


9 


13 


50rs 


AQst 


19 


17 


50a6 


406c 


9 


11 



QQ 6w + 12 3x 2_ 8x _ 9 

oo. 41. 



34. —^— 42. 

25w + 45 

35. 43. 

5 

16a; + 4 

36. 44. 

„„ 2s " 4 

37. 45. 

s 

„„ 7r ~ 8 

38. 46. 

r 

3r-5 

39. 47. 



40. 48. 

u lOur 



1. 


4s 3 

IT 


3. 


6c 

t 


5. 


s 3 
9£ 2 





X 2 




46 2 - 


-56- 


8 




6 2 




2a; 2 


— 3a; - 


-6 




a; 2 




6u 2 


— 5w - 


-2 




U 2 




12t 2 


+ 2t- 


-16 




I2t 2 




186 2 


: + 96 


- 15 




186 2 




4s 2 


+ 2s- 


10 




4s 2 




Ww 


2 + 12w - 2 



5*- $*> £»■ Answers ■*? -as -as 



7.il 

9c 2 



-I 



4 
11. 

3c 



7.3. SIMPLIFYING RATIONAL EXPRESSIONS 515 

Sx 



13. 

5yz 

17 z 
15. 

5 

17.-1 

2v 

5r 
19. 

2st 

44r + 15s 
21 " 72r 2 s 2 

15r + 34s 



23. 



72r 2 s 2 

3 i qq„,3 



25 28^ + 33y 



27. 



144y3 z 3 

15w 3 + 52v 3 
144v 3 uj 3 

44z - 45a; 



29. 

200xyz 



31. 


76c 


:-85a 


200abc 


33. 


2v- 


f 4 


35. 


5m 


+ 9 


37. 


2- 


4 
s 


39. 


3- 


5 
r 


41. 


3- 


8 9 

X X 2 


43. 


2- 


3 6 

X X 2 


45. 


1 + 


1 4 
6i ~ 3T 2 


47. 


1 + 


1 5 

Ys~ 2^ 



516 



CHAPTER 7. RATIONAL EXPRESSIONS 



7.4 Solving Rational Equations 

In Section 3 of Chapter 2, we showed that the most efficient way to solve an 
equation containing fractions was to first clear the fractions by multiplying 
both sides of the equation by the least common denominator. For example, 
given the equation 

1 1 _ 1 

2 X+ 3 = 4' 

we would first clear the fractions by multiplying both sides by 12. 



12 



ri i 




pi 


~x + - 


= 


— 


[2 3 




L 4 J 



- 12 



Qx + 4 = 3 



This procedure works equally well when the the denominators contain a vari- 
able. 



You Try It! 



Solve for x: 1 



6 



EXAMPLE 1. Solve for a;: 1 - - = — 



Solution: The common denominator is x 2 . We begin by clearing fractions, 
multiplying both sides of the equation by x 2 . 



1-1 = 1 

x x 2 





?,' 




r 3 1 


1 - 


_ 


— 






X _ 




X 2 



Original equation. 



Multiply both sides by x 



Now we use the distributive property. 



x 2 [1] - x 2 



Distribute x . 



Now we cancel common factors and simplify. 



2x = 3 



Cancel. Simplify. 



The resulting equation is nonlinear (x is raised to a power larger than 1 . Make 
one side zero, then factor. 



x z - 2x - 3 = 
(a:-3)(a; + l) = 



Nonlinear. Make one side zero. 
Factor. 



7.4. SOLVING RATIONAL EQUATIONS 



517 



Use the zero product property to complete the solution. Either the first factor 
is zero or the second factor is zero. 

a; — 3 = or £ + 1 = 
x = 3 x = — 1 

Hence, the solutions are x = —1 and x = 3. 

Check. Substitute —1 for x, then 3 for x in the original equation and simplify. 



1-2 * 

x x 2 



i-2 = A 

X X 2 



1 



(-1) (-1) 2 

1 + 2 = 3 
3 = 3 



1 



1 



(3) (S) 2 

2 _ 3 

~ 3 ~ 9 
1 1 

3 ~ 3 

Note that both result in true statements, showing that both x = — 1 and x = 3 

check in the original equation. Answer: 2, 4 



□ 



22 29 
EXAMPLE 2. Solve for x: 6- — = — 

x 2 x 

Solution: The common denominator is x 2 . 

22 _ 29 

72 



6 



x- 



:r 



x 



x 2 [6] 



22" 




[291 


6 TT 


=: 




X 2 




. x . 




[22" 




[291 


r." 





::i_: 







X 2 




X 



X 



6x 2 - 22 = 29x 



Original equation. 
Multiply both sides by x . 

Distribute x . 
Cancel and simplify. 



You Try It! 



7 
Solve for x: — 



This last equation is nonlinear. Make one side zero. 

6x 2 - 29x - 22 = 

The integer pair 4 and —33 has product ac = —132 and sum b = —29. Break 
up the middle term into a sum using this pair, then factor by grouping. 

6x 2 + Ax - 33x - 22 = 

2x(3x + 2)- ll(3x + 2) = 

(2x- ll)(3x + 2) = 



30 
x 



518 



CHAPTER 7. RATIONAL EXPRESSIONS 



Finally, use the zero product property to write: 



2x 



11 = 


or 


3a; + 2 = 


2a; = 11 




3a; = -2 


11 

X = Y 




2 



Answer: -1/4, -7/2 



Check: Let's check these solutions with our calculators. Enter 11/2, push the 
STO^ button, push the X,T,6>,n button and the ENTER key (see the calculator 
screen on the left in Figure 7.7). Next, enter the left-hand side of the equation 
as 6-22/X~2 and press ENTER. Enter the right-hand side of the equation as 
29/X and press ENTER. The results are the same (see the calculator screen on 
the left in Figure 7.7). This verifies that 11/2 is a solution of 6— 22/a; 2 = 29/a;. 

The calculator screen on the right in Figure 7.7 shows a similar check of 
the solution x = —2/3. 



ll/2-»X 

5.5 
6-22/X A 2 

5.272727273 
29/X 

5.272727273 



-2/3-»X 




-.6666666667 


6-22/X A 2 






-43.5 


29/X 






-43.5 



22 29 
Figure 7.7: Checking the solutions of 6 ~ = — ■ 



Solving Rational Equations with the Graphing Calculator 



□ 



Let's use the graphing calculator to solve an equation containing rational ex- 
pressions. 



You Try It! 



Solve the equation 
2 



5 
x 



12 



both algebraically and 
graphically, then compare 
your solutions. 



EXAMPLE 3. Consider the following equation: 

x x 

Solve the equation algebraically, then solve the equation graphically using your 
graphing calculator. Compare your solutions. 



7.4. SOLVING RATIONAL EQUATIONS 



519 



Algebraic solution: First, an algebraic approach. Multiply both sides of the 
equation by the common denominator x 2 . 



2-£ = A 

x x A 



x 2 \2\ 



9" 




' 5 " 


2 


— 




x _ 




X 1 


2 


"9" 




' 5 " 


X 


— 


^ 







X _ 




X 2 



2x z - 9x 



Original equation. 
Multiply both sides by x 2 

Distribute x . 
Cancel and simplify. 



The last equation is nonlinear. Make one side zero. 



2x z - 9a; 







Make one side zero. 



The integer pair —10 and 1 have product equaling ac = —10 and sum equaling 
b = —9. Break up the middle term using this pair, then factor by grouping. 



2x 2 - 10a; + x - 5 = 

2a;(x-5) + l(»-5) =0 

(2x+l)(cc-5) =0 



— 10a; + x = —9x. 
Factor by grouping. 
Factor out x — 5. 



Now use the zero product property to write: 



2a; + 1 = or 




x - 


-5 = 


2a; = -1 






x = 5 


1 
X = ~2 








re x = —1/2 and x = 


5. 







Hence, the solutions are x 

Graphical solution: We could load each side of the equation separately, then 
use the intersect utility to find where the graphs intersect. However, in this 
case, it's a bit easier to make one side of the equation zero, draw a single graph, 
then note where the graph crosses the a;-axis. 



9 5 

2-- = — 
x x A 

9 5 
2 - = 

x x z 



Original equation. 
Make one side zero. 



Load the left-hand side of the equation into Yl as 2-9/X-5/XA2 (see the image 
on the left in Figure 7.8), then select 6:ZStandard from the ZOOM menu to 
produce the image at the right in Figure 7.8. 
Next, the solutions of 

9 5 



2 

x x z 







520 



CHAPTER 7. RATIONAL EXPRESSIONS 



Ploti Plots Plots 
\ViB2-9/X-5^T2 

■xVs = 

■xVfi = 



\r 



Figure 7.8: Sketch the graph of y = 2 



9 
x 



are found by noting where the graph of y = 2 — 9/a: — 5/x 2 cross the a;-axis. 
Select 2 : zero from the CALC menu. Use the arrow keys to move the cursor 
to the left of the first x-intercept, then press ENTER to set the "Left bound." 
Next, move the cursor to the right of the first x-intercept, then press ENTER to 
set the "Right bound." Finally, leave the cursor where it is and press ENTER 
to set your "Guess." The calculator responds with the result shown in the 
figure on the left in Figure 7.9. 

Repeat the zero-finding procedure to capture the coordinates of the second 
x-intercept (see the image on the right in Figure 7.9). 







\ 

2*K0 
K=".£ 


V 



2*K0 




V=(i 



Figure 7.9: Locating the zeros of y = 2 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 7.10). 

• Place your WINDOW parameters at the end of each axis (see Figure 7.10). 

• Label the graph with its equation (see Figure 7.10). 

• Drop dashed vertical lines through each x-intercept. Shade and label 
the cc-values of the points where the dashed vertical line crosses the x- 
axis. These are the solutions of the equation 2 — 9/x — 5/x 2 = (see 
Figure 7.10). 



7.4. SOLVING RATIONAL EQUATIONS 



521 



y = 2 - 9/x - 5/x 2 




Figure 7.10: Reporting your graphical solution on your homework. 



Thus, the calculator is reporting that the solutions of 2 — 9/z — 5/x 2 = are 
x = —0.5 and x = 5, which match the algebraic solutions x = —1/2 and x = 5. 



Numerical Applications 

Let's apply what we've learned to an application. 



Answer: —4, 3/2 







: f 


K=-4 1 


v=o 







: t /~ 


K=1.E I 





□ 



You Try It! 



EXAMPLE 4. The sum of a number and its reciprocal is 41/20. Find the The sum of a number and its 

number. reciprocal is 53/14. Find the 

number . 
Solution: In the solution, we address each step of the Requirements for Word 

Problem Solutions. 



1. Set up a variable dictionary. Let x represent the unknown number. 

2. Set up an equation. If the unknown number is x, then its reciprocal is 
1/x. Thus, the "sum of a number and its reciprocal is 41/20" becomes: 



x + 



1 _ 41 
x ~ 20 



522 CHAPTER 7. RATIONAL EXPRESSIONS 

3. Solve the equation. Clear the fractions by multiplying both sides by 20a;, 
the least common denominator. 

1 41 

x H = — Model equation. 

x 20 

11 [41 
20a; x + - = — 20a; Multiply both sides by 20a;. 



1" 




[411 


x+ - 


= 





X _ 




L20J 




\Y 




[41 "1 


211.7: 


— 


= 







X _ 




L20J 



20a; \x] + 20a; - = — 20a; Distribute 20a;. 

20x + 20 = 41a; Cancel and simplify. 

The equation is nonlinear. Make one side zero. 

20x 2 - 41a; + 20 = Make one side zero. 

The integer pair —16 and —25 has product ac = 400 and sum 6 = —41. 
Break up the middle term in the last equation into a sum of like terms 
using this pair, then factor by grouping. 

20a; 2 - 16a; - 25a; + 20 = - 16a; - 25a; = -41a;. 

4a;(5a; — 4) — 5(5a; — 4) = Factor by grouping. 

(4a; - 5) (5a; - 4) = Factor out 5a; - 4. 

We can now use the zero product property to write: 



4a; - 5 = 


or 


5a; 


-4 = 


4a; = 5 






5a; = 4 


5 
X= 4 






4 
X= 5 



Answer: 2/7, 7/2 



4. Answer the question. There are two possible numbers, 5/4 and 4/5. 

5. Look back. The sum of the unknown number and its reciprocal is supposed 
to equal 41/20. The answer 5/4 has reciprocal 4/5. Their sum is: 

5 4 _ 16 25 

4 + 5 ~~ 20 + 20 
41 

~ 20 

Thus, 5/4 is a valid solution. 

The second answer 4/5 has reciprocal 5/4, so it is clear that their sum 
is also 41/20. Hence, 4/5 is also a valid solution. 



□ 



7.4. SOLVING RATIONAL EQUATIONS 523 



**>**■**> Exercises ■** * «•* 



In Exercises 1-8, solve the equation. 



1. 


26 

x = 11 + — 

a; 


2. 


„ 60 

x = 7H 


3. 


12 27 




x a; 2 


4. 


1 6 7 
x x z 







10 11 


5. 


1- 


X X 2 

20 96 


6. 


1- 


X X 2 

44 


7. 


x = 


= 7+ — 
a; 

99 


8. 


a; = 


= 2 + — 
a; 



In Exercises 9-16, solve the equation. 



9. 


12a; = 97 




a; 


10. 


7x = -19 




a: 


11. 


19 3 

20+ — = - 

x x z 


12. 


8 1 
33 - - = -s 

x x z 



13. 


11 
8x = 19 




X 


14. 


3 

28x = 25 




X 


15. 


40 + ^ = 1 
x x z 


16. 


11 1 

18 + — = ? 

x x z 



In Exercises 17-20, solve each equation algebraically, then use the calculator to check your solutions. 

17. 36x=-13 19. 14a; = 9 

x x 

10 20 

18. 9x = 43+— 20. 3x = 16 

x x 



In Exercises 21-24, solve the equation algebraically, then solve the equation using the graphing calcu- 
lator using the technique shown in Example 3. Report your solution using the Calculator Submission 
Guidelines demonstrated in Example 3. 

I 12 44 
21.1-- = — 23. 2x = 3 + — 

X X A X 

II 28 4 
22.1 + — = 24. 2x = 9-- 

xx 2 - X 



524 



CHAPTER 7. RATIONAL EXPRESSIONS 



25. The sum of a number and its reciprocal is 
5/2. Find the number. 

26. The sum of a number and its reciprocal is 
65/8. Find the number. 



27. The sum of a number and 8 times its recip- 
rocal is 17/3. Find all possible solutions. 

28. The sum of a number and 4 times its recip- 
rocal is 17/2. Find all possible solutions. 



j*- j*- ?*- Answers •** ■*$ ■** 



1. -2, 13 
3. 3, 9 
5. 11, -1 

7. -4, 11 

9. 8, 1/12 
11. -3/4, -1/5 
13. 11/8, 1 



15. -1/4, 1/10 
17. -1/9, -1/4 
19. 1/2, 1/7 
21. -3, 4 
23. -4, 11/2 
25. 2, 1/2 
27. 3, 8/3 



7.5. DIRECT AND INVERSE VARIATION 525 

7.5 Direct and Inverse Variation 

We start with the definition of the phrase "is proportional to." 

Proportional. We say that y is proportional to x if and only if 

y = kx, 

where k is a constant called the constant of proportionality. The phrase "y 
varies directly as x" is an equivalent way of saying "y is proportional to x." 

Here are a few examples that translate the phrase "is proportional to." 

• Given that d is proportional to t, we write d = kt, where k is a constant. 

• Given that y is proportional to the cube of x, we write y = kx 3 , where k 
is a constant. 

• Given that s is proportional to the square of t, we write s = kt 2 , where 
k is a constant. 

We are not restricted to always using the letter k for our constant of propor- 
tionality. 



You Try It! 



EXAMPLE 1. Given that y is proportional to x and the fact that y = 12 Given that y is proportional 
when x = 5, determine the constant of proportionality, then determine the to x and that y = 21 when 
value of y when x = 10. x = 9, determine the value of 

Solution: Given the fact the y is proportional to x, we know immediately 
that 

y = kx, 

where k is the proportionality constant. Because we are given that y = 12 
when x = 5, we can substitute 12 for y and 5 for x to determine k. 

y = kx y is proportional to x. 

12 = fc(5) Substitute 12 for y, 5 for x. 

12 

— = k Divide both sides by 5. 

5 

Next, substitute the constant of proportionality 12/5 for k in y = kx, then 
substitute 10 for x to determine y when x = 10. 

12 
y = — x Substitute 12/5 for k. 

5 

12 
y = — (10) Substitute 10 for x. 

5 

y = 24 Cancel and simplify. 

Answer: 63 

□ 



526 



CHAPTER 7. RATIONAL EXPRESSIONS 



You Try It! 



A ball is dropped from the 
edge of a cliff on a certain 
planet. The distance s the 
ball falls is proportional to 
the square of the time t that 
has passed since the ball's 
release. If the ball falls 50 
feet during the first 5 
seconds, how far does the 
ball fall in 8 seconds? 



EXAMPLE 2. A ball is dropped from a balloon floating above the surface 
of the earth. The distance s the ball falls is proportional to the square of the 
time t that has passed since the ball's release. If the ball falls 144 feet during 
the first 3 seconds, how far does the ball fall in 9 seconds? 

Solution: Given the fact the s is proportional to the square of t, we know 
immediately that 

s = kt 2 , 

where k is the proportionality constant. Because we are given that the ball 
falls 144 feet during the first 3 seconds, we can substitute 144 for s and 3 for t 
to determine the constant of proportionality. 



s = kt 2 
144 = fc(3) 2 
144 = 9fc 
16 = k 



s is proportional to the square of t. 
Substitute 144 for s, 3 for t. 
Simplify: 3 2 = 9. 
Divide both sides by 9. 



Next, substitute the constant of proportionality 16 for k in s = kt 2 , and then 
substitute 9 for t to determine the distance fallen when t = 9 seconds. 



I6r 

16(9) 2 
1296 



Substitute 16 for k. 
Substitute 9 for t. 
Simplify. 



Answer: 128 feet 



Thus, the ball falls 1,296 feet during the first 9 seconds. 



□ 



You Try It! 



If a 0.75 pound weight 
stretches a spring 5 inches, 
how far will a 1.2 pound 
weight stretch the spring? 



EXAMPLE 3. Tony and Paul are hanging weights on a spring in the physics 
lab. Each time a weight is hung, they measure the distance the spring stretches. 
They discover that the distance y that the spring stretches is proportional to 
the weight hung on the spring (Hooke's Law). If a 0.5 pound weight stretches 
the spring 3 inches, how far will a 0.75 pound weight stretch the spring? 

Solution: Let W represent the weight hung on the spring. Let y represent the 
distance the spring stretches. We're told that the distance y the spring stretches 
is proportional to the amount of weight W hung on the spring. Hence, we can 
write: 



y = kW 



y is proportional to W . 



7.5. DIRECT AND INVERSE VARIATION 527 

Substitute 3 for y, 0.5 for W, then solve for k. 

3 = fc(0.5) Substitute 3 for y, 0.5 for W. 

3 

= k Divide both sides by 0.5. 

0.5 

k = 6 Simplify. 

Substitute 6 for k in y = kW to produce: 

y = 6W Substitute 6 for k in y = kW . 

To determine the distance the spring will stretch when 0.75 pounds are hung 
on the spring, substitute 0.75 for W. 

y = 6(0.75) Substitute 0.75 for W. 

y = 4.5 Simplify. 

Thus, the spring will stretch 4.5 inches. Answer: 8 inches 

□ 



Inversely Proportional 

In Examples 1, 2, and 3, where one quantity was proportional to a second 
quantity, you may have noticed that when one quantity increased, the second 
quantity also increased. Vice-versa, when one quantity decreased, the second 
quantity also decreased. 

However, not all real-world situations follow this pattern. There are times 
when as one quantity increases, the related quantity decreases. For example, 
consider the situation where you increase the number of workers on a job and 
note that the time to finish the job decreases. This is an example of a quantity 
being inversely proportional to a second quantity. 

Inversely proportional. We say the y is inversely proportional to x if and 

only if 

k 

V = -i 
x 

where k is a constant called the constant of proportionality. The phrase "y 
varies inversely as x" is an equivalent way of saying "y in inversely proportional 
to x." 

Here are a few examples that translate the phrase "is inversely proportional 
to." 

• Given that d is inversely proportional to t, we write d = k/t, where k is 
a constant. 



528 



CHAPTER 7. RATIONAL EXPRESSIONS 



• Given that y is inversely proportional to the cube of x, we write y = k/x 3 , 
where fc is a constant. 

• Given that s is inversely proportional to the square oft, we write s = k/t 2 , 
where k is a constant. 



You Try It! 



We are not restricted to always using the letter k for our constant of propor- 
tionality. 



Given that y is inversely EXAMPLE 4. Given that y is inversely proportional to x and the fact that 

proportional to x and that y = 4 when x = 2, determine the constant of proportionality, then determine 

y = 5 when x = 8, determine the value of y when x = 4. 

the value of y when x = 10. . 

Solution: Given the fact the y is inversely proportional to a;, we know imme- 
diately that 

k 

V = -, 
x 

where k is the proportionality constant. Because we are given that y = 4 when 
x = 2, we can substitute 4 for ?/ and 2 for a; to determine k. 



k 

x 

k 



y 

4 

2 

8 = fc 



y is inversely proportional to x. 

Substitute 4 for y, 2 for x. 
Multiply both sides by 2. 



Substitute 8 for fc in y = k/x, then substitute 4 for x to determine y when 
x = 4. 

o 

y = — Substitue 8 for fc. 



y= 4 

y = 2 



Substitute 4 for x. 
Reduce. 



Answer: 4 



Note that as x increased from 2 to 4, y decreased from 4 to 2. 



□ 



You Try It! 



If the light intensity 4 feet 
from a light source is 2 
foot-candles, what is the 
intensity of the light 8 feet 
from the light source? 



EXAMPLE 5. The intensity I of light is inversely proportional to the square 
of the distance d from the light source. If the light intensity 5 feet from the 
light source is 3 foot-candles, what is the intensity of the light 15 feet from the 
light source? 



7.5. DIRECT AND INVERSE VARIATION 



529 



Solution: Given the fact that the intensity / of the light is inversely propor- 
tional to the square of the distance d from the light source, we know immedi- 
ately that 

7=1 

d 2 ' 

where k is the proportionality constant. Because we are given that the intensity 
is J = 3 foot-candles at d = 5 feet from the light source, we can substitute 3 
for / and 5 for d to determine k. 



I is inversely proportional to d 

Substitute 3 for /, 5 for d. 

Simplify. 

Multiply both sides by 25. 





k 


/ - 






d 2 




k 


3 = 






b 2 




k 


3 = 






2b 


75 = 


k 



Substitute 75 for k in I = k/d 2 , then substitute 15 for d to determine I when 
d= 15. 



75 

d 2 
75 

15 2 " 
75 

225 
1 

3 



Substitute 75 for k. 
Substitute 15 for d. 
Simplify. 
Reduce. 



Thus, the intensity of the light 15 feet from the light source is 1/3 foot-candle. 



Answer: 1/2 foot-candle 



□ 



You Try It! 



EXAMPLE 6. Suppose that the price per person for a camping experience is 
inversely proportional to the number of people who sign up for the experience. 
If 10 people sign up, the price per person is $350. What will be the price per 
person if 50 people sign up? 

Solution: Let p represent the price per person and let N be the number of 
people who sign up for the camping experience. Because we are told that the 
price per person is inversely proportional to the number of people who sign up 
for the camping experience, we can write: 



Suppose that the price per 
person for a tour is inversely 
proportional to the number 
of people who sign up for the 
tour. If 8 people sign up, the 
price per person is $70. 
What will be the price per 
person if 20 people sign up? 



P= N> 



Answer: $28 



530 CHAPTER 7. RATIONAL EXPRESSIONS 



where k is the proportionality constant. Because we are given that the price 
per person is $350 when 10 people sign up, we can substitute 350 for p and 10 
for N to determine k. 

k . . 

p is inversely proportional to N. 



p = 


N 


350 = 


k 
10 


3500 = 


k 



Substitute 350 for p, 10 for JV. 
Multiply both sides by 10. 

Substitute 3500 for k in p = k/N, then substitute 50 for N to determine p 
when N = 50. 

p = Substitute 3500 for k. 

N 

p = Substitute 50 for TV. 

1 50 

p = 70 Simplify. 

Thus, the price per person is $70 if 50 people sign up for the camping experi- 
ence. 

□ 



7.5. DIRECT AND INVERSE VARIATION 



531 



ti. ;». ;». 



Exercises 



■*j ■*'. •*; 



1. Given that s is proportional to £ and the 
fact that s = 632 when £ = 79, determine 
the value of s when £ = 50. 

2. Given that s is proportional to £ and the 
fact that s = 264 when £ = 66, determine 
the value of s when £ = 60. 

3. Given that s is proportional to the cube 
of £ and the fact that s = 1588867 when 
£ = 61, determine the value of s when 
£ = 63. 

4. Given that d is proportional to the cube 
of £ and the fact that d = 318028 when 
£ = 43, determine the value of d when 
£ = 76. 

5. Given that q is proportional to the square 
of c and the fact that q = 13448 when 
c = 82, determine the value of q when 
c = 29. 

6. Given that q is proportional to the square 
of c and the fact that q = 3125 when 
c = 25, determine the value of q when 
c = 87. 



7. Given that y is proportional to the square 
of x and the fact that y = 14700 when 
x = 70, determine the value of y when 
x = 45. 

8. Given that y is proportional to the square 
of x and the fact that y = 2028 when 
x = 26, determine the value of y when 
x = 79. 

9. Given that F is proportional to the cube 
of x and the fact that F = 214375 when 
x = 35, determine the value of F when 
x = 36. 

10. Given that d is proportional to the cube 
of t and the fact that d = 2465195 when 
t = 79, determine the value of d when 
£ = 45. 

11. Given that d is proportional to t and the 
fact that d = 496 when t = 62, determine 
the value of d when t = 60. 

12. Given that d is proportional to t and the 
fact that d = 405 when t = 45, determine 
the value of d when t = 65. 



13. Given that h is inversely proportional to 
x and the fact that h = 16 when x = 29, 
determine the value of h when x = 20. 

14. Given that y is inversely proportional to 
x and the fact that y = 23 when x = 15, 
determine the value of y when j; = 10. 

15. Given that q is inversely proportional to 
the square of c and the fact that q = 11 
when c = 9, determine the value of q when 
c = 3. 



16. Given that s is inversely proportional to 
the square of t and the fact that s = 11 
when t = 8, determine the value of s when 
t= 10. 

17. Given that F is inversely proportional to 
x and the fact that F = 19 when x = 22, 
determine the value of F when x = 16. 



18. Given that d is inversely proportional to 
t and the fact that d = 21 when £ = 16, 
determine the value of d when t = 24. 



532 



CHAPTER 7. RATIONAL EXPRESSIONS 



19. Given that y is inversely proportional to 
the square of x and the fact that y = 14 
when x = 4, determine the value of y 
when x = 10. 

20. Given that d is inversely proportional to 
the square of t and the fact that d = 21 
when t = 8, determine the value of ci when 
t = 12. 

21. Given that d is inversely proportional to 
the cube of t and the fact that d = 18 
when £ = 2, determine the value of ci when 
£ = 3. 



22. Given that g is inversely proportional to 
the cube of c and the fact that q = 10 
when c = 5, determine the value of q when 
c = 6. 

23. Given that q is inversely proportional to 
the cube of c and the fact that q = 16 
when c = 5, determine the value of <? when 
c = 6. 

24. Given that q is inversely proportional to 
the cube of c and the fact that q = 15 
when c = 6, determine the value of <? when 
c = 2. 



25. Joe and Mary are hanging weights on a 
spring in the physics lab. Each time a 
weight is hung, they measure the distance 
the spring stretches. They discover that 
the distance that the spring stretches is 
proportional to the weight hung on the 
spring. If a 2 pound weight stretches the 
spring 16 inches, how far will a 5 pound 
weight stretch the spring? 

26. Liz and Denzel are hanging weights on a 
spring in the physics lab. Each time a 
weight is hung, they measure the distance 
the spring stretches. They discover that 
the distance that the spring stretches is 
proportional to the weight hung on the 
spring. If a 5 pound weight stretches the 
spring 12.5 inches, how far will a 12 pound 
weight stretch the spring? 

27. The intensity / of light is inversely pro- 
portional to the square of the distance d 
from the light source. If the light inten- 
sity 4 feet from the light source is 20 foot- 
candles, what is the intensity of the light 



18 feet from the light source? 

28. The intensity / of light is inversely pro- 
portional to the square of the distance d 
from the light source. If the light inten- 
sity 5 feet from the light source is 10 foot- 
candles, what is the intensity of the light 
10 feet from the light source? 

29. Suppose that the price per person for a 
camping experience is inversely propor- 
tional to the number of people who sign 
up for the experience. If 18 people sign 
up, the price per person is $204. What 
will be the price per person if 35 people 
sign up? Round your answer to the near- 
est dollar. 

30. Suppose that the price per person for a 
camping experience is inversely propor- 
tional to the number of people who sign 
up for the experience. If 17 people sign 
up, the price per person is $213. What 
will be the price per person if 27 people 
sign up? Round your answer to the near- 
est dollar. 



7.5. DIRECT AND INVERSE VARIATION 



533 



m- j*- j*- Answers •*$ •** •** 



1. 400 

3. 1750329 

5. 1682 

7. 6075 

9. 233280 
11. 480 
13. 116/5 
15. 99 



17. 209/8 

19. 56/25 

21. 16/3 

23. 250/27 

25. 40 inches 

27. 1.0 foot-candles 

29. $105 



Chapter 8 

Quadratic Functions 



In this chapter, we will be solving equations using the inverse operations of 
squaring and square roots (radicals). We will be using the familiar Pythagorean 
theorem to solve problems with right triangles. In addition, we will learn how 
use the technique of "completing the square" to develop the quadratic formula 
to help solve quadratic equations that are not factorable by methods of previous 
chapters. 

Four thousand years ago, the Babylonians were solving selected quadratic 
equations. The Greek mathematician Pythagoras did geometrical studies of 
which the Pythagorean theorem for right triangles is well known even in modern 
times. 

In circa 250 BC Diophantes of Alexandria and in 750 AD al Khwarizmis 
of Arabia also wrote quadratic equations with solutions. These earlier math- 
ematicians limited their work to positive rational numbers, but in medieval 
times the Hindu Brahmagupta included solutions using positive and negative 
numbers and also included irrational numbers as well. 

In the 1600's, Frenchman Rene Descartes wrote a geometric proof of the 
quadratic formula which simplified the previous information. His generation 
of mathematicians formulated the symbols which make it easier for us to solve 
problems with radicals and higher powers. 



535 



536 



CHAPTER 8. QUADRATIC FUNCTIONS 



8.1 Introduction to Radical Notation 

We know how to square a number. For example: 

• 5 2 = 25 

• (-5) 2 = 25 

Taking the square root of a number is the opposite of squaring. 

• The nonnegative square root of 25 is 5. 

• The negative square root of 25 is —5. 

Thus, when searching for a square root of a number, we are searching for 
number whose square is equal to our number. 

EXAMPLE 1. Find the square roots of 81. 

Solution: We are looking for a number whose square is 81. 

• Because 9 2 = 81, the nonnegative square root of 81 is 9. 

• Because (— 9) 2 = 81, the negative square root of 81 is —9. 

Hence, 81 has two square roots, —9 and 9. 

□ 



You Try It! 



Find the square roots of 64. 



Answer: 8 and 



You Try It! 



Find the square roots of 100. EXAMPLE 2. Find the square roots of 0. 

Solution: We are looking for a number whose square is 0. 

• Because 2 = 0, the nonnegative square root of is 0. 

No other number squared will equal zero. Hence, zero has exactly one square 
Answer: 10 and —10 root, namely zero. 

□ 



You Try It! 



Find the square roots of 
-25. 



EXAMPLE 3. Find the square roots of -36. 



.1. INTRODUCTION TO RADICAL NOTATION 



537 



Solution: We are looking for a number whose square is —36. However, every 
time you square a real number, the result is never negative. Hence, —36 has 
no real square roots. 1 



Answer: no real square roots 



The introductions in Examples 1, 2, and 3 lead to the following definition. 



□ 



Defining the square roots of a number. The solutions of x 2 = a are called 
square roots of a. 

Case: a > 0. The equation x 2 = a has two real solutions, namely x = ±-y/a. 

• The notation y/a calls for the nonegative square root. 

• The notation —y/a calls for the negative square root. 

Case: a = 0. The equation x 2 = has exactly one solution, namely x = 0. 
Case: a < 0. The equation x 2 = a has no real solutions. 



EXAMPLE 4. Solve x 2 = 9 for x, then simplify your answers. 

Solution: Because the right-hand side of x 2 = 9 is positive, the equation has 
two solutions. 



You Try It! 



Solve x 2 = 16 for x, then 
simplify your answers. 



9 
±V9 



Original equation. 

Two answers: — v9 and v9 



To simplify these answers, we need to understand the following facts: 

• V9 calls for the nonnegative square root of 9. Because (3) 2 = 9, the 
nonnegative square root of 9 is 3. Hence, V9 = 3. 

• — V9 calls for the negative square root of 9. Because (— 3) 2 = 9, the 
negative square root of 9 is —3. Hence, — V9 = —3. 



Thus, the solutions of x 2 
or x = 3." 



9 are x = ±3, which is equivalent to saying "x 



Answer: 4, 



□ 



1 When we say that —36 has no real square roots, we mean there are no real numbers 
that are square roots of —36. The reason we emphasize the word real in this situation is 
the fact that —36 does have two square roots that are elements of the complex numbers, a 
set of numbers that are usually introduced in advanced courses such as college algebra or 
trigonometry. 



538 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



Solve x 2 = 49 for x, then 
simplify the answers. 



Answer: 7, 



EXAMPLE 5. Solve x 2 = for x, then simplify the answer. 
Solution: There is only one number whose square equals 0, namely 0. 



x 2 = 
x = 



Original equation. 
One answer: (0) = 0. 



Thus, the only solution of x 2 = is x = 0. Consequently, the nonnegative 
square root of zero is zero. Hence, V0 = 0. 

□ 



You Try It! 



Solve 



-9 for x, then 



simplify the answers. 
Answer: no real solutions 



EXAMPLE 6. Solve 



-4 for x, then simplify the answer. 



Solution: You cannot square a real number and get a negative result. Hence, 



-4 has no real solutions. Therefore, 



-4 is not a real number. 



□ 



You Try It! 



Simplify: -\/l44 



EXAMPLE 7. Simplify each of the following: 

a) y/Ul b) -v/225 



c) V^wo 



d) 



/324 



Solution: Remember, the notation y/a calls for the nonnegative square root 
of a, while the notation —y/a calls for the negative square root of a. 

a) Because ll 2 = 121, the nonnegative square root of 121 is 11. Thus: 



'121 = 11 



b) Because (— 15) 2 = 225, the negative square root of 225 is —15. Thus: 



7 225 



-15 



Answer: —12 



c) You cannot square a real number and get —100. Therefore, V - 100 is not 
a real number. 

d) Because (— 18) 2 = 324, the negative square root of 324 is —18. Thus: 



7 324 



-18 



□ 



.1. INTRODUCTION TO RADICAL NOTATION 



539 



Squaring "undoes" taking the square root. 



Squaring square roots. If a > 0, then both —yfa and ^fa are solutions of 



a. Consequently, if we substitute each of them into the equation x 



we get: 



(-V5) s 



and 



W' 



EXAMPLE 8. Simplify each of the following expressions: 

a) (VE) 2 b) (-V7) 2 

Solution: We'll handle each case carefully. 



c) (V^TT) 2 



a) Because V5 is a solution of x 2 = 5, if we square V5, we should get 5. 

(A) 2 = 5 

b) Because — \fl~ is a solution of x 2 = 7, if we square —y/7, we should get 7. 



(-V7) 2 



7 



You Try It! 



Simplify: (-v 7 ^) 5 



c) Because x 2 = — 11 has no real answers, \J — 11 is not a real number. Ad- 
vanced courses such as college algebra or trigonometry will introduce the 
complex number system and show how to handle this expression. 



Answer: 21 



□ 



Using the Graphing Calculator 

Up to this point, the equation x 2 = a has involved perfect squares. For exam- 
ple, if we start with x 2 = 25, then the solutions are x = ±v25. Because 25 is 
a perfect square, we can simplify further, arriving at x = ±5. 

However, the right-hand side of x 2 = a does not have to be a perfect square. 
For example, the equation x 2 = 7 has two real solutions, x = ±V7. Because 7 
is not a perfect square, we cannot simplify further. In the next example, we'll 
use the graphing calculator to compare this algebraic solution with a graphical 
solution and hopefully provide some assurance that — v7 and v7 are perfectly 



valid solutions of ; 



7. 



540 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



Solve the equation x 2 = 5 
both algebraically and 
graphically, then compare 
your answers. 



EXAMPLE 9. Use the graphing calculator to solve x 2 = 7. Then solve the 
equation algebraically and compare answers. 

Solution: Enter each side of the equation x 2 = 7 in the Y= menu (see 
Figure 8.1), then select 6:ZStandard to produce the image in Figure 8.1. 



Plotl PlotE Plots 




\ 


E / 


WiBX A 2 




\ 


: / 


-■■ViB? 




\ 


: / 


Wj=I 




\ 


J. 






We = 








^Vs= 








\V? = 









Figure 8.1: Sketch each side of x 2 = 7 . 

Use the 5: intersect utility on the CALC menu to find the points of intersec- 
tion. Press ENTER in response to "First curve," press ENTER in response to 
"Second curve," then use the arrow keys to move the cursor closer to the point 
of intersection on the left than the one on the right. Press ENTER in response 
to "Guess." This will produce the point of intersection shown in image on the 
left in Figure 8.2. Repeat the procedure to find the point of intersection in the 
image on the right in Figure 8.2. 



V 


[ / 


\ 


\l 


Intersection 
K=-£.fiH£7£i 


•{=? 



\ 


[ J. 


\ 


\l 


Intersection 
K=2.fiH£7£13 


•{=? 



Figure 8.2: Finding points of intersection. 



The approximate solutions of x 2 = 7 are x fs —2.645751 and x « 2.6457513. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 8.3). 

• Place your WINDOW parameters at the end of each axis (see Figure 8.3). 

• Label each graph with its equation (see Figure 8.3). 

• Drop dashed vertical lines through each point of intersection. Shade and 
label the £- values of the points where the dashed vertical line crosses the 
x-axis. These are the solutions of the equation x 2 = 7 (see Figure 8.3). 



.1. INTRODUCTION TO RADICAL NOTATION 



541 




Figure 8.3: Reporting your graphical solution on your homework. 



Now we solve the equation algebraically. 

2 n 

x = 7 
x = ±V7 

At this point, the question is: "Do these algebraic solutions match the graphical 
solutions in Figure 8.3?" Let's use our calculator to compare results. 

Locate the square root symbol \J on the calculator case above the x 2 
key in the leftmost column on the calculator keyboard. Note that we will have 
to use the 2nd key to access this operator. Enter — <J~ (7) and press ENTER. 
Then enter y/~ (7) and press ENTER. The results are shown in Figure 8.4. 



Answer: — v5, Vo 



-JX7S 

-2.645751311 



4"C7) 



2.645751311 



Figure 8.4: Computing — \fl and v7. 

Thus, -y/7 ~ -2.645751311 and V7 w 2.645751311. Note how these closely 
match the graphical approximations in Figure 8.3. 



-■J"C5> 

-2.236S&7977 



4"<5) 



2.236067977 



V 


1/ 


\ 


:/ 


Intersection 
K=-2.23fi0fiH 


V=E 



\ 


!/ 


\ 


:/ 


InttFS«ction 
X=2.23G0fiB 


V=E 



D 



542 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



List of Squares 



n 


n 2 








1 


1 


2 


4 


3 


9 


4 


16 


5 


25 


6 


36 


7 


49 


8 


64 


9 


81 


10 


100 


11 


121 


12 


144 


13 


169 


14 


196 


15 


225 


16 


256 


17 


289 


18 


324 


19 


361 


20 


400 


21 


441 


22 


484 


23 


529 


24 


576 


25 


625 



EXAMPLE 10. Use the graphing calculator to solve x 2 
the equation algebraically and compare answers. 



-5. Then solve 



Solution: Enter each side of the equation x 2 = —5 in the Y= menu (see 
Figure 8.5), then select 6:ZStandard to produce the image in Figure 8.5. 



Plotl 


PlOtE 


plots 


WiBX A 2 




-■■ViB- 


■5 




Wj = 






^Vh= 






We = 






vVs = 






\V? = 







Figure 8.5: Sketch each side of x 2 = —5. 

Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 8.6). 

• Place your WINDOW parameters at the end of each axis (see Figure 8.6). 

• Label each graph with its equation (see Figure 8.6). 

Because there are no points of intersection, the graph in Figure 8.6 informs us 
that the equation x 2 = — 5 has no real solutions. 
Now we solve the equation algebraically. 



Table 8.1: However, you cannot square a real number and get a negative answer. Hence, 

the equation x 2 = — 5 has no real solutions. This agrees completely with the 
graph in Figure 8.6. 

□ 



Approximating Square Roots 

The squares in the "List of Squares" shown in Table 8.1 are called perfect 
squares. Each is the square of a whole number. But not all numbers are 
perfect squares. For example, in the case of v24, there is no whole number 
whose square is equal to 24. However, this does not prevent V24 from being a 
perfectly good number. 



.1. INTRODUCTION TO RADICAL NOTATION 



543 



v 1 

y = x 





\ 10 


, 


10 


10 






y = - 




-10 


■ 



Figure 8.6: Reporting your graphical solution on your homework. 



We can use the "List of Squares" to find decimal approximations when the 
radicand is not a perfect square. 



You Try It! 



EXAMPLE 11. Estimate V24 by guessing. Use a calculator to find a more Estimate: V83 
accurate result and compare this result with your guess. 

Solution. From the "List of Squares," note that 24 lies betwen 16 and 25, so 
V24 will lie between 4 and 5, with V24 much closer to 5 than it is to 4. 



16 



+ 



24 

4- 



25 
I > 



4.8 



Let's guess 

As a check, let's square 4.; 



24 



(4.8) 2 = (4.8)(4.8) = 23.04 



Not quite 24! Clearly, V24 must be a little bit bigger than 4.8. 

Let's use a scientific calculator to get a better approximation. From our 
calculator, using the square root button, we find 



24 « 4.89897948557. 



544 CHAPTER 8. QUADRATIC FUNCTIONS 

Even though this is better than our estimate of 4.8, it is still only an ap- 
proximation. Our calculator was only capable of providing 11 decimal places. 
However, the exact decimal representation of v24 is an infinite decimal that 
never terminates and never establishes a pattern of repetition. 

Just for fun, here is a decimal approximation of V24 that is accurate to 
1000 places, courtesy of http: //www. wolf ramalpha.com/. 

4.8989794855663561963945681494117827839318949613133402568653851 

3450192075491463005307971886620928046963718920245322837824971773 

09196755146832515679024745571056578254950553531424952602105418235 

40446962621357973381707264886705091208067617617878749171135693149 

44872260828854054043234840367660016317961567602617940145738798726 

16743161888016008874773750983290293078782900240894528962666325870 

21889483627026570990088932343453262850995296636249008023132090729 

18018687172335863967331332533818263813071727532210516312358732472 

35822058934417670915102576710597966482011173804100128309322482347 

06798820862115985796934679065105574720836593103436607820735600767 

24633259464660565809954782094852720141025275395093777354012819859 

11851434656929005776183028851492605205905926474151050068455119830 

90852562596006129344159884850604575685241068135895720093193879959 

87119508123342717309306912496416512553772738561882612744867017729 

60314496926744648947590909762887695867274018394820295570465751182 

126319692156620734019070649453 

If you were to multiply this number by itself (square the number) , you would 
get a number that is extremely close to 24, but it would not be exactly 24. 
Answer: 9.1 There would still be a little discrepancy. 

□ 



.1. INTRODUCTION TO RADICAL NOTATION 



545 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



1. List all real square roots of —400. 

2. List all real square roots of 64. 

3. List all real square roots of —25. 

4. List all real square roots of 81. 

5. List all real square roots of 49. 



6. List all real square roots of —100. 

7. List all real square roots of 324. 

8. List all real square roots of 36. 

9. List all real square roots of —225. 
10. List all real square roots of 0. 



11. List all real solutions of x 2 

12. List all real solutions of x 2 

13. List all real solutions of x 2 



-225. 
-25. 



361. 



14. List all real solutions of x 2 = 256. 

15. List all real solutions of x 2 = —400. 



16. List all real solutions of x 2 = 0. 

17. List all real solutions of x 2 = 169. 



18. List all real solutions of ; 



-100. 



19. List all real solutions of x 2 = 625. 

20. List all real solutions of x 2 = 324. 



In Exercises 21-30, simplify each of the given expressions 
^64 



21. v 

22. -V^529 

23. - v /z 256 



24. V^529 

25. -v/361 



26. ^^361 

27. -x/T00 

28. -VT96 

29. \/441 

30. V49 



In Exercises 31-38, simplify each of the given expressions. 

31. (-VTT) 2 35. (-^/29) i 

32. (-\ / 3l) 2 36. (-y/M)' 

33. (V59) 2 37. (779) 2 

34. (V43) 2 38. (^3) 2 



546 



CHAPTER 8. QUADRATIC FUNCTIONS 



In Exercises 39-42, for each of the given equations, first use the 5:intersect utility on the CALC menu 
of the graphing calculator to determine the solutions. Follow the Calculator Submission Guidelines, 
as demonstrated in Example 9 in reporting the solution on your homework paper. Second, solve the 
equation algebraically, then use your calculator to find approximations of your answers and compare 
this second set with the first set of answers. 



39. x 2 = 37 

40. x 2 = 32 



41. x 2 = 11 

42. x 2 = 42 



;*•;*■ t*- 



Answers 



•*;•*; •*! 



1. There are no real square roots. 

3. There are no real square roots. 

5. -7, 7 

7. -18, 18 

9. There are no real square roots. 
11. There are no real solutions. 
13. -19, 19 

15. There are no real solutions. 
17. -13, 13 
19. -25, 25 
21. 8 

23. The expression is not a real number. 
25. -19 
27. -10 
29. 21 
31. 17 
33. 59 



35. 29 



37. 79 



39. 



x 50 > 

\ 1 
\l 


i / 


l\ 
1 \ 
1 \ 
1 \ 
1 \ 


1 \ 

/ i 
/ i 
/ i 
J? 


10 i 

-6.082763 
i 
i 
i 
i 
i 
1 -50 ^ 


i 10 

6.082763 
i 
i 
i 
i 
i 
■ i 



37 



Answers: ±V37 w ±6.082763 



41. 





\ i 

\ i 

\ i 

\ i 


30 


. 
i / 
i / 
i / 
i / 
i / 




l\ 
1 \ 
l X. 


A 
/ 1 


10 


i 
i 




i 
i 


10 



y = 11 



-3.316625 3.316625 



-30 



Answers: ±VH « ±3.316625 



,2. SIMPLIFYING RADICAL EXPRESSIONS 



547 



8.2 Simplifying Radical Expressions 

Let's begin by comparing two mathematical expressions 
V^v^ =3-4 



12 



V9 • 16 = Vui 
= 12 



Note that both vWHI and V9 ■ 16 equal 12. Hence, v^v 7 !^ = V9 ■ 16. Let's 
look at another example. 



VZV9 = 2-3 
= 6 



\AT9 = V36 
= 6 



Note that both v4v9 and v4 • 9 equal 6. Hence, v4V9 = v4 ■ 9. It appears 
that a pattern is forming, namely: 

\favb = vab 



Let's try an example on our calculator. First enter V2v3, then enter \/2 • 3 
(see Figure 8.7). Note that they produce the same result. Therefore, V2v3 = 
y/2T3. 



■J"t2>*J"C3> 

2.449489743 



■J"C2*3> 

2.4494S9743 



Figure 8.7: Note that V2\/3 = V2 _ 3. 

The above discussion leads us to the following result. 

Multiplication property of radicals. If a > and 6 > 0, then: 

\fa\fb = yah 



You Try It! 



EXAMPLE 1. Simplify each of the following expressions as much as possible: Simplify: V2v8 

a) ypSs/Yi b) VT2V3 c) V2VT3 

Solution: In each case, use the property y/avb = \/ab. That is, multiply the 
two numbers under the square root sign, placing the product under a single 
square root. 



548 



CHAPTER 8. QUADRATIC FUNCTIONS 



Answer: 4 



VsVll = V3- 11 b) VuVs = V12-3 c) \/2\/l3 = V2 ■ 13 
= V33 = V36 = ^26 

= 6 



□ 



Simple Radical Form 

We can also use the property y/avb = \/ab in reverse to factor out a perfect 
square. For example: 

VoO = v25v2 Factor out a perfect square. 

= 5\/2 Simplify: %/25 = 5. 

The expression 5v2 is said to be in simple radical form. Like reducing a fraction 
to lowest terms, you should always look to factor out a perfect square when 
possible. 

Simple radical form. If possible, always factor out a perfect square. 



You Try It! 



Place v 12 in simple radical 
form. 



EXAMPLE 2. Place \/8 in simple radical form. 

Solution: From V8, we can factor out a perfect square, in this case V4. 

V 8 = v 4v 2 Factor out a perfect square. 

= 2\f2 Simplify: y/l = 2. 



Answer: 2V3 



You Try It! 



□ 

Sometimes, after factoring out a perfect square, you can still factor out 
another perfect square. 



Place V200 in simple radical EXAMPLE 3. Place v 72 in simple radical form. 

form. . , — / — 

Solution: From V72, we can factor out a perfect square, in this case v 9. 

Factor out a perfect square. 
Simplify: V9 = 3. 



72 = V9V8 

:3V^ 



,2. SIMPLIFYING RADICAL EXPRESSIONS 



549 



However, from v8 we can factor out another perfect square, in this case V4. 

= 3\/4V2 
= 3-2-V2 

= 6\/2 



Factor out another perfect square. 
Simplify: y/l = 2. 
Multiply: 3-2 = 6. 



Alternate solution. We can simplify the process by noting that we can factor 
out V36 from V72 to start the process. 

V 36v 2 Factor out a perfect square. 

6v2 Simplify: v 36 = 6. 



72 



Although the second solution is more efficient, the first solution is still math- 
ematically correct. The point to make here is that we must continue to factor 
out a perfect square whenever possible. Our answer is not in simple radical 
form until we can no longer factor out a perfect square. Answer: 



10\/2 



The Pythagorean Theorem 

An angle that measures 90 degrees is called a right angle. If one of the angles 
of a triangle is a right angle, then the triangle is called a right triangle. It is 
traditional to mark the right angle with a little square (see Figure 8.8). 



□ 




A b C 

Figure 8.8: Right triangle AABC has a right angle at vertex C . 



Right triangle terminology. 






• The longest side of the right triangle, 


the side directly opposite the right 


angle, is called the hypotenuse of the 


right triangle. 




• The remaining two sides of the right triangle are called the legs 


of the 


right triangle. 







550 



CHAPTER 8. QUADRATIC FUNCTIONS 



Proof of the Pythagorean Theorem 

Each side of the square in Figure 8.9 has been divided into two segments, one 
of length a, the other of length b. 




Figure 8.9: Proving the Pythagorean Theorem. 



We can find the total area of the square by squaring any one of the sides of the 
square. 

A = (a + b) Square a side to find area. 

A = a + 2ab + b Squaring a binomial pattern. 

Thus, the total area of the square is A = a 2 + 2ab + b 2 . 

A second approach to finding the area of the square is to sum the areas of 
the geometric parts that make up the square. We have four congruent right 
triangles, shaded in light red, with base a and height 6. The area of each of 
these triangles is found by taking one-half times the base times the height; i.e., 
the area of each triangles is (l/2)ab. In the interior, we have a smaller square 
with side c. Its area is found by squaring its side; i.e., the area of the smaller 
square is c 2 . 

The total area of the square is the sum of its parts, one smaller square and 
four congruent triangles. That is: 



4 | -ab 



2ab 



Adding the area of the interior square 
and the area of four right triangles. 

Simplify: 4((l/2)a6) = 2ab. 



The two expressions, a 2 + 2ab + b 2 and c 2 + 2ab, both represent the total area 
of the large square. Hence, they must be equal to one another. 

a + 2ab + b = c + 2ab Each side of this equation represents 

the area of the large square. 

a + b = c Subtract 2ab from both sides. 



,2. SIMPLIFYING RADICAL EXPRESSIONS 



551 



The last equation, a 2 



c 2 , is called the Pythagorean Theorem. 



The Pythagorean Theorem. If a and b are the legs of a right triangle and 
c is its hypotenuse, then: 



2 i 1,2 2 
a + b = c 



We say "The sum of the squares of the legs of a right triangle equals the square 
of its hypotenuse." 



Good hint. Note that the hypotenuse sits by itself on one side of the equation 
a 2 + b 2 = c 2 . The legs of the hypotenuse are on the other side. 
Let's put the Pythagorean Theorem to work. 



You Try It! 



EXAMPLE 4. Find the length of the missing side of the right triangle shown Find the missing side of the 
below. right triangle shown below. 




Solution: First, write out the Pythagorean Theorem, then substitute the 
given values in the appropriate places. 



a + 

2 i fo\2 



(4) 2 + (3) 

16 + 9 

25 



Pythagorean Theorem. 
Substitute: 4 for a, 3 for b. 
Square: (4) 2 = 16, (3) 2 = 9. 
Add: 16 + 9 = 25. 




The equation c 2 = 25 has two real solutions, c = —5 and c = 5. However, in 
this situation, c represents the length of the hypotenuse and must be a positive 
number. Hence: 



c = 5 Nonnegative square root. 

Thus, the length of the hypotenuse is 5. 



Answer: 13 



□ 



552 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



An isosceles right triangle 
has a hypotenuse of length 
10. Find the lengths of the 
legs. 



EXAMPLE 5. An isosceles right triangle has a hypotenuse of length 8. Find 
the lengths of the legs. 

Solution: In general, an isosceles triangle is a triangle with two equal sides. 
In this case, an isosceles right triangle has two equal legs. We'll let x represent 
the length of each leg. 




Use the Pythagorean Theorem, substituting x for each leg and 8 for the hy- 
potenuse. 



2 i t.2 2 

a + b = c 
x 2 + x 2 = 8 2 
2x 2 = 64 
x 2 = 32 



Pythagorean Theorem. 
Substitute: x for a, x for b, 8 for a 
Combine like terms: x + x = 2x. 
Divide both sides by 2. 



The equation x 2 = 32 has two real solutions, x = — v32 and x = V32- How- 
ever, in this situation, x represents the length of each leg and must be a positive 
number. Hence: 



x = \/32 



Nonnegative square root. 



Remember, your final answer must be in simple radical form. We must factor 
out a perfect square when possible. 



Answer: Each leg has 
length 5\/2. 



x = V16v2 Factor out a perfect square. 

x = iV2 Simplify: Vl6 = 4. 



Thus, the length of each leg is 4V2- 



□ 



,2. SIMPLIFYING RADICAL EXPRESSIONS 



553 



Applications 

Let's try a word problem. 



EXAMPLE 6. A ladder 20 feet long leans against the garage wall. If the 
base of the ladder is 8 feet from the garage wall, how high up the garage wall 
does the ladder reach? Find an exact answer, then use your calculator to round 
your answer to the nearest tenth of a foot. 

Solution: As always, we obey the Requirements for Word Problem Solutions. 

1. Set up a variable dictionary. We'll create a well-marked diagram for this 
purpose, letting h represent the distance between the base of the garage 
wall and the upper tip of the ladder. 




You Try It! 



A ladder 15 feet long leans 
against a wall. If the base of 
the ladder is 6 feet from the 
wall, how high up the wall 
does the ladder reach? Use 
your calculator to round 
your answer to the nearest 
tenth of a foot. 



2. Set up an equation. Using the Pythagorean Theorem, we can write: 



64 + /i 2 



3. Solve the equation. 



h = 20 Pythagorean Theorem. 

400 Square: 8 2 = 64 and 20 2 = 400. 



h = 336 Subtract 64 from both sides. 

h = v336 h will be the nonnegative square root. 

h = v 16 v 21 Factor out a perfect square. 

h = 4V^T Simplify: VIE = 4. 



4. Answer the question. The ladder reaches 4v21 feet up the wall. Using a 
calculator, this is about 18.3 feet, rounded to the nearest tenth of a foot. 



554 



CHAPTER 8. QUADRATIC FUNCTIONS 



5. Look back. Understand that when we use 18.3 ft, an approximation, our 
solution will only check approximately. 




18.3 ft 



Using the Pythagorean Theorem: 

8 2 + 18. 3 2 = 20 2 

64 + 334.89 = 400 

398.89 = 400 

The approximation is not perfect, but it seems close enough to accept 
this solution. 



Answer: 13.7 feet 



□ 



,2. SIMPLIFYING RADICAL EXPRESSIONS 



555 



ti. ;». ;». 



Exercises 



■*j ■*-: •*; 



In Exercises 1-6, simplify the given expression, writing your answer using a single square root symbol. 
Check the result with your graphing calculator. 



1. V^Vu 

2. y/2\/l 

3. VT7V2 



4. \/Ey/Tl 

5. v^VTf 

6. VT7V3 



In Exercises 7-26, convert each of the given expressions to simple radical form. 



7. 


V56 


8. 


^45 


9. 


v/99 


10. 


v 7 ^ 


11. 


VT50 


12. 


v^ 


13. 


\/40 


14. 


\/m 


15. 


a/28 


16. 


7175 



17. 


V153 


18. 


^/T25 


19. 


a/50 


20. 


a/88 


21. 


V^8 


22. 


vTff 


23. 


V44 


24. 


v^o 


25. 


a/ToI 


26. 


V27 



In Exercises 27-34, find the length of the missing side of the right triangle. Your final answer must be 
in simple radical form. 



27. 




28. 




556 



CHAPTER 8. QUADRATIC FUNCTIONS 




32. 





33. 




34. 




35. In the figure below, a right triangle is in- 
scribed in a semicircle. What is the area 
of the shaded region? 

C 




36. In the figure below, a right triangle is in- 
scribed in a semicircle. What is the area 
of the shaded region? 




37. The longest leg of a right triangle is 10 feet 
longer than twice the length of its shorter 



leg. The hypotenuse is 4 feet longer than 
three times the length of the shorter leg. 
Find the lengths of all three sides of the 
right triangle. 

38. The longest leg of a right triangle is 2 feet 
longer than twice the length of its shorter 
leg. The hypotenuse is 3 feet longer than 
twice the length of the shorter leg. Find 
the lengths of all three sides of the right 
triangle. 

39. A ladder 19 feet long leans against the 
garage wall. If the base of the ladder is 5 
feet from the garage wall, how high up the 
garage wall does the ladder reach? Use 
your calculator to round your answer to 
the nearest tenth of a foot. 

40. A ladder 19 feet long leans against the 
garage wall. If the base of the ladder is 6 
feet from the garage wall, how high up the 
garage wall does the ladder reach? Use 
your calculator to round your answer to 
the nearest tenth of a foot. 



:.2. SIMPLIFYING RADICAL EXPRESSIONS 



557 



m- j*- j*- Answers •*$ •** •** 



1. \/65 

3. ^34 

5. ^85 

7. 2v/l4 

9. 3\/TT 

11. 5\/6 

13. 2i/l0 

15. 2^/7 

17. 3VT7 

19. 5\/2 



21. 


3^2 


23. 


2\/II 


25. 


2v / 26 


27. 


2^15 


29. 


2^154 


31. 


2^37 


33. 


2^74 


35. 


25 k 

7T — 

8 


37. 


7, 24, 25 


39. 


18.3 feet 



558 



CHAPTER 8. QUADRATIC FUNCTIONS 



8.3 Completing the Square 

In Introduction to Radical Notation on page 536, we showed how to solve equa- 
tions such as x 2 = 9 both algebraically and graphically. 



±3 



y = 9 




■10 * 



Note that when we take the square root of both sides of this equation, there 
are two answers, one negative and one positive. 

A perfect square is nice, but not required. Indeed, we may even have to 
factor out a perfect square to put our final answer in simple form. 



x 



x = ±V8 

x = ±VIV2 
x = ±2^2 




You Try It! 



Readers should use their calculators to check that — 2y2 ss —2.8284 and 2y2 : 
2.8284. 

Now, let's extend this solution technique to a broader class of equations. 



Solve for a;: (x + 6) 2 = 16. EXAMPLE 1. Solve for x: (x - 4) 2 



9. 



Solution: Much like the solutions of x 2 = 9 are x = ±3, we use a similar 
approach on (x — 4) 2 = 9 to obtain: 



(a;-4) 2 = 9 
x - 4 = ±3 



Original equation. 

There are two square roots. 



8.3. COMPLETING THE SQUARE 559 

To complete the solution, add 4 to both sides of the equation. 

x = 4 ± 3 Add 3 to both sides. 

Note that this means that there are two answers, namely: 

£ = 4-3 or £ = 4 + 3 
x = 1 x = 7 

Check: Check each solution by substituting it into the original equation. 
Substitute 1 for x: Substitute 7 for x: 

(a;-4) 2 = 9 (x - 4) 2 = 9 

(l-4) 2 = 9 (7-4) 2 = 9 

(-3) 2 = 9 (3) 2 = 9 

Because the last statement in each check is a true statement, both x = 1 and 

x = 7 are valid solutions of (x — 4) 2 = 9. Answer: —2, —10 

□ 

In Example 1, the right-hand side of the equation (x — 4) 2 = 9 was a perfect 
square. However, this is not required, as the next example will show. 



You Try It! 



EXAMPLE 2. Solve for x: (x + 5) 2 = 7. Solve for x: (x - 4) : 

Solution: Using the same technique as in Example 1, we obtain: 
(x + 5) =7 Original equation. 

x + 5 = ±v 7 There are two square roots. 

To complete the solution, subtract 5 from both sides of the equation. 

x = — 5 ± v7 Subtract 5 from both sides. 

Note that this means that there are two answers, namely: 
x = — 5 — v7 or x = — 5 + v7 

Check: Check each solution by substituting it into the original equation. 
Substitute —5 — V7 for x: Substitute —5 + V7 for x: 

( x + 5) 2 = 7 ( x + 5) 2 = 7 

((-5 - y/7) + 5) 2 = 7 ((-5 + y/f) + 5) 2 = 7 

(-V7) 2 = 7 (V7) 2 = 7 



560 



CHAPTER 8. QUADRATIC FUNCTIONS 



Answer: 4 + \/5, 4 - \/5 



You Try It! 



Solve for x: (x + 7) 2 = 18. 



Because the last statement in each check is a true statement, both x = — 5— v7 
and x = —5 + V7 are valid solutions of (x + 5) 2 = 7. 

□ 

Sometimes you will have to factor out a perfect square to put your answer 
in simple form. 



EXAMPLE 3. Solve for x: (x + 4) 2 = 20. 

Solution: Using the same technique as in Example 1, we obtain: 

(x + 4) = 20 Original equation. 

x + 4 = ±v20 There are two square roots. 

x + 4 = ±v 4v 5 Factor out a perfect square. 

x + 4 = ±2^5 Simplify: y/i=2. 

To complete the solution, subtract 4 from both sides of the equation. 

X = —4 ± 2v 5 Subtract 4 from both sides. 

Note that this means that there are two answers, namely: 
£ = -4-2^ or ^ = -4 + 2^ 



Check: Although it is possible to check the exact answers, let's use our cal- 
culator instead. First, store —4 — 2y5 in X. Next, enter the left-hand side of 
the equation (x + 4) 2 = 20 (see image on the left in Figure 8.10). Note that 
(x + 4) 2 simplifies to 20, showing that —4 — 2\/H is a solution of (x + 4) 2 = 20. 
In similar fashion, the solution —4 + 2v5 also checks in (x + 4) 2 = 20 (see 
image on the right in Figure 8.10). 



-4-2*J"<5)-*X 

-S. 472135955 
(X+4> A 2 

20 



-4+2*J"<5)-*X 

.472135955 
(X+4> A 2 

20 



Figure 8.10: Checking —4 — 2V5 and —4 + 2V5 in the equation (x + 4) z 



20. 



Answer: 

-7 + 3\/2, -7-3a/2 



□ 



1.3. COMPLETING THE SQUARE 



561 



Perfect Square Trinomials Revisited 

Recall the squaring a binomial shortcut. 



Squaring a 


binomial. If a and b 


are any real numbers, 


then: 




(a±b) 2 


= a 2 ± 2ab + b 2 




That is, you 


square the first term, 


take the product of the first and second 


terms and double the result, then square the third term. 





Reminder examples: 

(x + 3) 2 =x 2 + 2(»(3) + 3 2 
= x 2 + 6 s + 9 



(x-8) 2 = x 2 - 2(s)(8) + 8 2 
= x 2 - 16a; + 64 



Because factoring is "unmultiplying," it is a simple matter to reverse the mul- 
tiplication process and factor these perfect square trinomials. 



x 2 + 6s + 9 = (s + 3) 2 



16x + 64= (x 



Note how in each case we simply take the square root of the first and last terms. 



EXAMPLE 4. Factor each of the following trinomials: 

a) x 2 - Ylx + 36 b) x 2 + 10s + 25 c) x 2 - 34x + 289 

Solution: Whenever the first and last terms of a trinomial are perfect squares, 
we should suspect that we have a perfect square trinomial. 

a) The first and third terms of x 2 — Ylx + 36 are perfect squares. Hence, we 
take their square roots and try: 

x 2 - Ylx + 36= (x-6) 2 

Note that 2(cc)(6) = 12a;, which is the middle term on the left. The solution 
checks. 

b) The first and third terms of a: 2 + 10a: + 25 are perfect squares. Hence, we 
take their square roots and try: 

a; 2 + 10s + 25 = (x + 5) 2 

Note that 2(s)(5) = 10s, which is the middle term on the left. The solution 
checks. 



You Try It! 



Factor: x 2 + 30s + 225 



562 



CHAPTER 8. QUADRATIC FUNCTIONS 



Answer: (a; + 15)^ 



c) The first and third terms of x 2 — 34a; + 289 are perfect squares. Hence, we 
take their square roots and try: 

x 2 - 34a; + 289 = (x - 17) 2 

Note that 2(a;)(17) = 34a;, which is the middle term on the left. The 
solution checks. 



□ 



Completing the Square 

In this section we start with the binomial x 2 + bx and ask the question "What 
constant value should we add to x 2 + bx so that the resulting trinomial is a 
perfect square trinomial?" The answer lies in this procedure. 



Completing the square. To calculate the constant required to make 


x 2 + bx 


a perfect square trinomial: 








1. Take one- half of the coefficient of x: 


b 
2 






fb\ 
2. Square the result of step one: — 1 


2 _b 2 
4 






3. Add the result of step two to x 2 + bx: 


2 , b2 
x + bx H 

4 






If you follow this process, the result will be 


a perfect square 


trinomial which 


will factor as follows: 

2 , b 2 ( 

x + bx H = \ x 

4 V 


n 2 







You Try It! 



Given x 2 + 16a;, complete EXAMPLE 5. Given x 2 + 12x, complete the square to create a perfect 

the square to create a perfect square trinomial. 

square trinomial. o„,„ iS „„. n„ „_„ Ji , „„„,: tl ,J 



Solution: Compare x + 12a; with x + bx and note that 

1. Take one-half of 12: 6 

2. Square the result of step one: 6 2 = 36 



12. 



3. Add the result of step two to x 2 + Ylx: 



12a; + 36 



1.3. COMPLETING THE SQUARE 



563 



Check: Note that the first and last terms of x 2 + 12a; + 36 are perfect squares. 
Take the square roots of the first and last terms and factor as follows: 

a; 2 + 12a; + 36 = (x + 6) 2 

Note that 2(a;)(6) = 12a;, so the middle term checks. 



Answer: 
x 2 + 16a; ■ 



64 = (x 



□ 



You Try It! 



EXAMPLE 6. Given a; 2 — 3a;, complete the square to create a perfect square Given a; 2 — 5a;, complete the 



trinomial. 

Solution: Compare x 2 — 3a; with a; 2 + bx and note that 

3 
1. Take one-half of —3: 



2. Square the result of step one: 



3\ 2 _ 9 
l) ~ 4 



3. Add the result of step two to x 2 — 3a;: x 2 — 3x + 



9 



square to create a perfect 
square trinomial. 



Check: Note that the first and last terms of x 2 — 3a; + 9/4 are perfect squares. 
Take the square roots of the first and last terms and factor as follows: 



9 



x -3x + -=[x-- 
4 V 2 



Note that 2(x) (3/2) = 3a;, so the middle term checks. 



Answer: 

a; 2 - 5a; + 10/4 = (x - h/2f 



□ 



Solving Equations by Completing the Square 

Consider the following nonlinear equation. 

a; 2 = 2a; + 2 

The standard approach is to make one side zero and factor. 

a; 2 - 2a; - 2 = 

However, one quickly realizes that there is no integer pair whose product is 
ac = — 2 and whose sum is b = — 2. So, what does one do in this situation? 
The answer is "Complete the square." 



564 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



Use completing the square to 
help solve x 2 = 3 — 6x. 



EXAMPLE 7. Use completing the square to help solve x 2 = 2x + 2. 

Solution: First, move 2x to the left-hand side of the equation, keeping the 
constant 2 on the right-hand side of the equation. 

x 2 - 2x = 2 

On the left, take one-half of the coefficient of x: (l/2)(— 2) = — 1. Square the 
result: (— l) 2 = 1. Add this result to both sides of the equation. 

x 2 - 2x + 1 = 2 + 1 
x 2 - 2x + 1 = 3 

We can now factor the left-hand side as a perfect square trinomial. 

(x -l) 2 = 3 

Now, as in Examples 1, 2, and 3, we can take the square root of both sides of 
the equation. Remember, there are two square roots. 



x 



1 = ±^3 



Finally, add 1 to both sides of the equation. 

x = 1±V3 

Thus, the equation x 2 = 2x + 2 has two answers, x = 1 — V3 and x = 1 + v3- 

Check: Let's use the calculator to check the solutions. First, store 1 — v3 in X 
(see the image on the left in Figure 8.11). Then enter the left- and right-hand 
sides of the equation x 2 = 2x + 2 and compare the results (see the image on 
the left in Figure 8.11). In similar fashion, check the second answer 1 + V3 
(see the image on the right in Figure 8.11). 



1-J"<3^X 




.7320508076 


X A 2 






.5358983849 


2+X+2 






.5358983849 



1+J"C3^X 

2.732050808 

7.464101615 
2+X+2 

7.464101615 



Figure 8.11: Checking 1 — \/3 and 1 + \/3 in the equation x 2 = 2x + 2. 



Answer: 

-3 + 2^, -3-2^3 



In both cases, note that the left- and right-hand sides oi x 2 = 2a;+2 produce the 
same result. Hence, both 1 — v3 and 1 + V3 are valid solutions of x 2 = 2x + 2. 



□ 



3.3. COMPLETING THE SQUARE 



565 



EXAMPLE 8. Solve the equation x 2 — 8x — 12 = 0, both algebraically and 
graphically. Compare your answer from each method. 

Algebraic solution: First, move the constant 12 to the right-hand side of 
the equation. 



You Try It! 



Solve the equation 
x 2 + 6x + 3 = both 
algebraically and graphically, 
then compare your answers. 



8a; - 12 

x 2 -8x 




12 



Original equation. 
Add 12 to both sides. 



Take half of the coefficient of x: (l/2)(- 
add 16 to both sides of the equation. 



-4. Square: (-4) 2 = 16. Now 



x 2 - 8x + 16 = 12 + 16 
(aj - 4) 2 = 28 
x-4 = ±^28 



Add 16 to both sides. 
Factor left-hand side. 
There are two square roots. 



Note that the answer is not in simple radical form. 



x - 4 = ±yfWl 
x - 4 = ±2^7 
x = A±2Vl 



Factor out a perfect square. 
Simplify: VI = 2. 
Add 4 to both sides. 



Graphical solution: Enter the equation y = x 2 — 8x — 12 in Yl of the 
Y= menu (see the first image in Figure 8.12). After some experimentation, we 
settled on the WINDOW parameters shown in the middle image of Figure 8.12. 
Once you've entered these WINDOW parameters, push the GRAPH button to 
produce the rightmost image in Figure 8.12. 



Plotl PlotE Plots 
WiBX A 2-8*X-12 
-■■Vi = 

Wj = 
\Vh= 

We = 

\V?= 



WINDOW 
Xnin= _ 5 
Xnax=15 
Xscl=l 
Ymin=-50 
Vnax=5@ 
Vscl=16 

l¥.res=W 




Figure 8.12: Drawing the graph of y 



8x- 12. 



We're looking for solutions of x 2 — 8x — 12 = 0, so we need to locate where 
the graph of y = x 2 — 8x — 12 intercepts the x-axis. That is, we need to find 
the zeros of y = x 2 — 8x — 12. Select 2:zero from the CALC menu, move the 
cursor slightly to the left of the first ^-intercept and press ENTER in response 
to "Left bound." Move the cursor slightly to the right of the first x-intercept 



566 



CHAPTER 8. QUADRATIC FUNCTIONS 



and press ENTER in response to "Right bound." Leave the cursor where it sits 
and press ENTER in response to "Guess." The calculator responds by finding 
the ^-coordinate of the x-intercept, as shown in the first image in Figure 8.13. 



Repeat the process to find the second ^-intercept of y 
in the second image in Figure 8.13. 



8a;— 12 shown 




K= -1.291503 V=0 




Z*K* 
K=9.29i£0£fi V=0 



Figure 8.13: Calculating the zeros of y 



8x- 12. 



Reporting the solution on your homework: Duplicate the image in your 
calculator's viewing window on your homework page. Use a ruler to draw all 
lines, but freehand any curves. 

• Label the horizontal and vertical axes with x and y, respectively (see 
Figure 8.14). 

• Place your WINDOW parameters at the end of each axis (see Figure 8.14). 

• Label the graph with its equation (see Figure 8.14). 

• Drop dashed vertical lines through each ir-intercept. Shade and label the 
a;- values of the points where the dashed vertical line crosses the x-axis. 
These are the solutions of the equation x 2 — 8x — 12 = (see Figure 8.14). 



2 -8x- 12 




-1.291503 



9.2915026 

>• x 



Figure 8.14: Reporting your graphical solution on your homework. 



3.3. COMPLETING THE SQUARE 



567 



Thus, the graphing calculator reports that the solutions of x 2 — 8a; — 12 = Answer: —3 — \f&, —3 + V& 
are x w -1.291503 and x « 9.2915026. 

Comparing exact and calculator approximations. How well do the 
graphing calculator solutions compare with the exact solutions, x = 4 — 2v7 
and x = 4 + 2V7? After entering each in the calculator (see Figure 8.15), the 
comparison is excellent! 



-3-J"C6> 

-5.449489743 
-3+J"C6> 

-.5505102572 



4-2* JX7) 

-1.291502622 



4+2* JX7) 

9.291502622 




K=-£.HH9H9 V=-iE"12 



Figure 8.15: Approximating exact solutions x = 4 — 2v7 and a; = 4 + 2V7- 





1 


K= -.££05103 V=0 





D 



568 CHAPTER 8. QUADRATIC FUNCTIONS 



**. s*. j*- Exercises «•< •** * 

In Exercises 1-8, find all real solutions of the given equation. Place your final answers in simple radical 
form. 

5. x 2 = -16 

6. x 2 = -104 

7. x 2 = 124 

8. x 2 = 148 



In Exercises 9-12, find all real solutions of the given equation. Place your final answers in simple radical 
form. 

9. (x + 19) 2 = 36 11. (x + 14) 2 = 100 

10. (x - 4) 2 = 400 12. (x- 15) 2 = 100 



1. 


x 2 = 84 


2. 


x 2 = 88 


3. 


x 2 = 68 


4. 


x 2 = 112 



In Exercises 13-18, square each of the following binomials. 

13. (x + 23) 2 16. (x-7) 2 

14. (x-5) 2 17. (x-25) 2 

15. (x + 11) 2 18. (x + 4) 2 



In Exercises 19-24, factor each of the following trinomials. 

19. x 2 + 24x + 144 22. x 2 + 8x + 16 

20. x 2 - 16x + 64 23. x 2 - 20x + 100 

21. x 2 -34x + 289 24. x 2 + 16x + 64 



In Exercises 25-36, for each expression, complete the square to form a perfect square trinomial. Check 
your answer by factoring your result. Be sure to check your middle term. 

25. x 2 - 20x 27. x 2 - 6x 

26. x 2 - Wx 28. x 2 - 40x 



8.3. COMPLETING THE SQUARE 569 

29. x 2 + 20a; 33. x 2 + 15x 

30. a; 2 + 26a; 34. x 2 + 25a; 

31. a; 2 + 7a; 35. a; 2 - 5a; 

32. a; 2 + 19a; 36. a; 2 - 3a; 

In Exercises 37-52, find all real solutions, if any, of the given equation. Place your final answers in 
simple radical form. 

37. a; 2 = 18a; - 18 45. a; 2 = 16a; - 8 

38. a; 2 = 12a; - 18 46. a; 2 = 10a; - 5 

39. a; 2 = 16a; - 16 47. a; 2 = -18a; - 18 

40. a; 2 = 12a; - 4 48. a; 2 = -lOx - 17 

41. x 2 = -16a: - 4 49. a; 2 = -16x - 20 

42. a; 2 = -12x- 12 50. x 2 = -16a; - 12 

43. a; 2 = 18x - 9 51. x 2 = -18x-l 

44. a; 2 = 16a; - 10 52. x 2 = -12a; - 8 



In Exercises 53-56, solve the given equation algebraically, stating your final answers in simple radical 
form. Next, use the graphing calculator to solve the equation, following the technique outlined in 
Example 8. Use the Calculator Submission Guidelines, as demonstrated in Example 8, when reporting 
the solution on your homework. Compare the solutions determined by the two methods. 

53. a; 2 - 2x - 17 = 55. x 2 - 6x - 3 = 

54. x 2 - 4x - 14 = 56. x 2 - 4x - 16 = 



**■ s*< j*- Answers •*$ •** •** 

1. ±2^/2T 9. -25, -13 

3. ±2\/l7 11. -24, -4 

5. No real solutions 13. x 2 + 46a; + 529 

7. ±2^ 15. x 2 + 22x + 121 



570 CHAPTER 8. QUADRATIC FUNCTIONS 

17. x 2 - 50a; + 625 37.9-3^7,9 + 3^7 

19. (x + 12) 2 39.8-4^3,8 + 4^/3 

21. (x-17) 2 41. -8-2a/15, -8 + 2^15 

23. (x-10) 2 43. 9 - 6\/2, 9 + 6\/2 

25. x 2 - 20a; + 100 45.8-2714,8 + 2^/14 

27. a; 2 -6a; + 9 47. -9 - 3^7, -9 + 3\/7 

29. a; 2 + 20a; + 100 49. -8 - 2\/II, -8 + 2vTI 

31. a; 2 + 7a; + 49/4 51. -9 - 4^5, -9 + 4^5 

33. a; 2 + 15a; + 225/4 53. 1 - 3\/2, 1 + 3\/2 

35. a; 2 - 5a; + 25/4 55. 3 - 2i/3, 3 + 2^3 



8.4. THE QUADRATIC FORMULA 571 

8.4 The Quadratic Formula 

We first start with the definition of a quadratic equation. 

Quadratic equation. A second degree polynomial equation of the form 

ax 2 + bx + c = 0, 
where a, 6, and c are any real numbers, is called a quadratic equation in x. 

The goal of this section is to develop a formulaic shortcut that will provide 
exact solutions of the quadratic equation ax 2 + bx + c = 0. We start by moving 
the constant term to the other side of the equation. 

ax + bx + c = Quadratic equation. 

ax + bx = —c Subtract c from both sides. 

In preparation for completing the square, we next divide both sides of the 
equation by a. 



2 b 

x H — x = 


c 


— 


a 


a 



Divide both sides by a. 

Now we complete the square. Take one-half of the coefficient of x, then square 
the result. 

16 6 . . / 6 \ 2 6 2 



— ■ — = — when squared gives — = — r 

2 a 2a H B \2a 4a? 



We now add 6 /(4a ) to both sides of the equation. 



6 6 2 '■- 

a Aa 2 a 4a' 2 



x 2 + - x H = 1 Add 6 2 /(4a 2 ) to both sides. 



On the left, we factor the perfect square trinomial. On the right, we make 
equivalent fractions with a common denominator. 

c 4a b 2 

x -\ I = ■ 1 On the left, factor. On the right, 

2a J a 4a 4a 2 .' . ' 

create equivalent tractions with 

a common denominator. 

6\ 2 _ 4ac b 2 
Ya) ~~4a~ 2 + 4a 2 



x + — ) = — ^— j + -t— o Multiply numerators and denominators. 



6 2 - 4ac 

x H = — -= — Add fractions. 

2a J 4a 2 



572 



CHAPTER 8. QUADRATIC FUNCTIONS 



When we take the square root, there are two answers. 



b lb 2 - Aac 



Two square roots. 



When you take the square root of a fraction, you take the square root of both 
the numerator and denominator. 



b \]b 2 - Aac 

2a ^4^2 



b Vb 2 - Aac 

x-\ = ± 

2a 2a 



Simplify: v4a 2 = 2a. 



b \/b 2 — Aac 

x = ± Subtract 6/ (2a) from both sides. 

2a 2a 

Because both fractions have the same denominator, we can add and subtract 
numerators and put the answer over the common denominator. 



You Try It! 



Solve for x: 

x 2 - 8x + 12 = 



-b± Vo 2 - 4ac 
2a 



The quadratic formula. The equation aa; 2 + bx + c 
equation. Its solutions are given by 


= is called a quadratic 


—b± Vo 2 - 4ac 
la 

called the quadratic formula. 



Whew! Fortunately, the result is a lot easier to apply than it is to develop! 
Let's try some examples. 



EXAMPLE 1. Solve for x: x 2 - 4x - 5 = 0. 

Solution: The integer pair 1,-5 has product ac 
Hence, this trinomial factors. 



-5 and sum b 



x — Ax- 5 
(s + l)(a;-5) 



Now we can use the zero product property to write: 





x + 1 
x 



or 



1 



£-5 = 
x = 5 



8.4. THE QUADRATIC FORMULA 573 

Thus, the solutions are x = — 1 and x = 5. 

Now, let's give the quadratic formula a try. First, we must compare our 
equation with the quadratic equation, then determine the values of a, o, and c. 

ax + bx + c = 
x 2 - 4x - 5 = 

Comparing equations, we see that a = 1, b = —4, and c = —5. We will now 
plug these numbers into the quadratic formula. First, replace each occurrence 
of a, 6, and c in the quadratic formula with open parentheses. 



x = Z^gZJ^ The quadratic formula . 

2a 



-( )±y/( )2-4( )( ) _ , , , ... 

x = ; — ; Replace a, o, and c with 

2 ' 

open parentheses. 

Now we can substitute: 1 for a, —4 for 6, and —5 for c. 



* = -M) ±V(-4) a -4(1)^5) Substitute: x for flj _ 4 for ft| 

and —5 for c. 



a; = Simplify. Exponent first, then 

multiplication. 

4± v/36 

x = — Add: 16 + 20 = 36. 

x = — - — Simplify: V36 = 6. 

Note that because of the "plus or minus" symbol, we have two answers. 

4-6 4+6 

x = or x = 

2 2 

-2 10 

x ~ ~2~ x ~ T 

x = — 1 x = 5 

Note that these answers match the answers found using the ac-test to factor 

the trinomial. Answer: 2, 6 



□ 



574 



CHAPTER 8. QUADRATIC FUNCTIONS 



You Try It! 



Solve for x: x 2 + 7x = 10 



EXAMPLE 2. Solve for x: x 2 = 5x + 7. 

Solution: The equation is nonlinear, make one side zero. 



x* = 5x + 7 



5a; - 7 = 



Original equation. 

Nonlinear. Make one side zero. 



Compare x 2 — 5x — 7 = with ax 2 + bx + c = and note that a = 1, b = —5, 
and c = —7. Replace each occurrence of a, b, and c with open parentheses to 
prepare the quadratic formula for substitution. 



-b± \/b 2 - Aac 

2o 
-( )±V( ) 2 ~4( )( ) 
2( ) 



The quadratic formula. 

Replace a, b, and c with 
open parentheses. 



-(-5)± V / (-5) 2 -4(l)(-7) 



Substitute 1 for a, —5 for b, and —7 for c. 

Substitute: a = 1, b = — 5, c = — 7. 
Exponents and multiplication first. 
Simplify. 



2(1) 



5 ± V25 + 28 



5± \/53 



Answer: 

(-7 + V89)/2, (-7- 



89)/2 



Check: Use the calculator to check each solution (see Figure 8.16). Note that 
in storing (5 — v53)/2 in X, we must surround the numerator in parentheses. 



<5-J"<53))/2+X 

-1.140054945 
X A 2 

1.299725277 
5+X+7 

1.299725277 



<5+JX53))/2+X 




6. 


140054945 


X*2 








37 


.70027472 


5+X+7 






37 


.70027472 


I 







Figure 8.16: Check (5 - V53)/2 and (5 + V53)/2. 

In each image in Figure 8.16, after storing the solution in X, note that the left- 
and right-hand sides of the original equation x 2 = 5x + 7 produce the same 
number, verifying that our solutions are correct. 

□ 



8.4. THE QUADRATIC FORMULA 575 

In addition to placing all square roots into simple radical form, sometimes 
you need to reduce your answer to lowest terms. 



You Try It! 



EXAMPLE 3. Solve for x: 7.x 2 - lOx + 1 = 0. Solve for x: 3x 2 + 8x + 2 = 

Solution: Compare 7a; 2 — lOx + 1 = with ax 2 + bx + c = and note that 
a = 7, b = —10, and c = 1. Replace each occurrence of a, b, and c with open 
parentheses to prepare the quadratic formula for substitution. 



— The quadratic formula. 



2a 

-( )±v/( ) 2 "4( )( ) 



Replace a, 6, and c with 



2( ) 

open parentheses. 



Substitute 7 for a, —10 for 6, and 1 for c. 



-(-io)±v/(-io) 2 - 

2(7) 


-4(7)(1) 


Substitute: 7 for a, 
— 10 for 6, and 1 for c 


10 ± V100-28 

14 

10±\/72 
x = 


Exponent, then multiplication 
Simplify. 



14 

In this case, note that we can factor out a perfect square, namely v36. 
10 ± V36V2 



Vl2 = \/36V2. 
14 

10 ±6^ 



Simplify: v 36 = 6 

Finally, notice that both numerator and denominator are divisible by 2. 

10±6v/2 

2 

x = Divide numerator and 

14 

~n denominator by 2. 

10 6A/2 

2 2 

x = Distribute the 2. 

14 



2 
5±3a/2 



Simplify. 



576 



CHAPTER 8. QUADRATIC FUNCTIONS 



Answer: 

(-4 + VlO)/3, (-4- 



10)/3 



Alternate simplification. Rather than dividing numerator and denominator 
by 2, some prefer to factor and cancel, as follows. 



10 ±6^ 
14 

2(5 ±3^2) 

2(7) 
2(5 ± 3^2) 

W) 
5±3y / 2 

7 



Original answer. 
Factor out a 2. 

Cancel. 

Simplify. 



Note that we get the same answer using this technique. 



□ 



You Try It! 



An object is launched 
vertically and its height y 
(in feet) above ground level 
is given by the equation 
y = i60 + 96i- 16i 2 , 
where t is the time (in 
seconds) that has passed 
since its launch. How much 
time must pass after the 
launch before the object 
returns to ground level? 



EXAMPLE 4. An object is launched vertically and its height y (in feet) 
above ground level is given by the equation y = 320 + 192t — 16f 2 , where t is the 
time (in seconds) that has passed since its launch. How much time must pass 
after the launch before the object returns to ground level? After placing the 
answer in simple form and reducing, use your calculator to round the answer 
to the nearest tenth of a second. 

Solution: When the object returns to ground level, its height y above ground 
level is y = feet. To find the time when this occurs, substitute y = in the 
formula y = 320 + 192i - 16£ 2 and solve for t. 



y = 320 + 192* - 16r 
= 320+192t- 16£ 2 

Each of the coefficients is divisible by —16. 



= t-12t- 20 



Original equation. 
Set y = 0. 



Divide both sides by —16. 



Compare t 2 - lit - 20 = with at 2 + bt + c = and note that a = 1, b = -12, 
and c = —20. Replace each occurrence of a, &, and c with open parentheses to 
prepare the quadratic formula for substitution. Note that we are solving for t 
this time, not x. 



-b ± Vb 2 - 4ac 
2a 



-( )±V( ) 2 ~4( )( ) 
2( ) 



The quadratic formula. 

Replace a, b, and c with 
open parentheses. 



•A. THE QUADRATIC FORMULA 



577 



Substitute 1 for 



-12 for b, and -20 for 



t = rs Substitute: 1 for a, 



I, 



2(1) 



12 ± V144 + 80 



-12 for b, and -20 for c 
Exponent, then multiplication. 

Simplify. 



The answer is not in simple form, as we can factor out vl6 
12± v / 16\ / 14 



I 



t 



12±4v / 14 



V224 = V16V14. 
Simplify: VT6 = 4 



Use the distributive property to divide both terms in the numerator by 2. 



12 4\/li 
t= —± — — 
2 2 

t = 6±2^ 



Divide both terms by 2. 
Simplify. 



Thus, we have two solutions, t = 6 — 2V14 and t = 6 + 2V14. Use your 
calculator to find decimal approximations, then round to the nearest tenth. 

i« -1.5, 13.5 

The negative time is irrelevant, so to the nearest tenth of a second, it takes the 
object approximately 13.5 seconds to return to ground level. 



6-2*J"04) 

-1.4S33 14774 
6+2*J"04) 

13.48331477 



Answer: 

3 + v!9 w 7.4 seconds 



□ 



You Try It! 



EXAMPLE 5. Arnie gets on his bike at noon and begins to ride due north 
at a constant rate of 12 miles per hour. At 1:00 pm, Barbara gets on her bike 
at the same starting point and begins to ride due east at a constant rate of 8 
miles per hour. At what time of the day will they be 50 miles apart (as the 
crow flies)? Don't worry about simple form, just report the time of day, correct 
to the nearest minute. 

Solution: At the moment they are 50 miles apart, let t represent the time that 
Arnie has been riding since noon. Because Barbara started at 1 pm, she has 
been riding for one hour less than Arnie. So, let t — 1 represent the numbers 
of hours that Barbara has been riding at the moment they are 50 miles apart. 



At 6:00 am, a freight train 
passes through Sagebrush 
Junction heading west at 40 
miles per hour. At 8:00 am, 
a passenger train passes 
through the junction heading 
south at 60 miles per hour. 
At what time of the day, 
correct to the nearest 
minute, will the two trains 
be 180 miles apart? 



578 



CHAPTER 8. QUADRATIC FUNCTIONS 




8(«-l) 

Figure 8.17: 50 miles apart. 



Now, if Arnie has been riding at a constant rate of 12 miles per hour for * 
hours, then he has traveled a distance of 12* miles. Because Barbara has been 
riding at a constant rate of 8 miles per hour for * — 1 hours, she has traveled a 
distance of 8(* — 1) miles. 

The distance and direction traveled by Arnie and Barbara are marked in 
Figure 8.17. Note that we have a right triangle, so the sides of the triangle 
must satisfy the Pythagorean Theorem. That is, 



(12t) 2 + [8{t - l)} 2 = 
Distribute the 8. 

(12i) 2 + (8t- 8) 2 = 
Square each term. Use (a — by 



144r + 64i 2 
208* 2 



128* + 64 : 

128* + 64 



50^ 



5(T 



2500 
2500 



Use the Pythagorean Theorem. 

Distribute the 8. 
2ab + b 2 to expand (8* - 8) 2 . 



Square each term. 
Simplify: 144* 2 + 64* 2 



208r 



The resulting equation is nonlinear. Make one side equal to zero. 



208r - 128* - 2436 = 
52* 2 - 32* - 609 = 



Subtract 2500 from both sides. 
Divide both sides by 4. 



Compare 52i 2 - 32* - 609 = with at 2 + bt + c = and note that a = 52, 
b = —32, and c = —609. Replace each occurrence of a, b, and c with open 
parentheses to prepare the quadratic formula for substitution. Note that we 
are solving for * this time, not x. 



-b± yfb 2 - 4ac 
2a 



-( )±V( ) 2 ~4( )( ) 
2( ) 



The quadratic formula. 

Replace a, 6, and c with 
open parentheses. 



Substitute 52 for a, —32 for b, and —609 for c 
-(-32) ± y(-32) 2 -4(52)(-609) 



t 



2(52) 



32 ± V1024+ 126672 
104 



* 



32 ± V127696 
104 



Substitute: 52 for a, 
-32 for 6, and -609 for c 

Exponent, then multiplication. 
Simplify. 



Now, as the request is for an approximate time, we won't bother with simple 
form and reduction, but proceed immediately to the calculator to approximate 



•A. THE QUADRATIC FORMULA 



579 



{32-JX1 27696) VI 
04 

-3. 12S32472 
C32+J"a27696>Vl 
04 

3.743709336 



Figure 8.18: Approximate time that Arnie has been riding. 



this last result (see Figure 8.18). Thus, Arnie has been riding for approxi- 
mately 3.743709336 hours. To change the fractional part 0.743709336 hours to 
minutes, multiply by 60 min/hr. 



0.743709336 hr = 0.743709336 hr x 



60min 



hr 



44.62256016 min 



Rounding to the nearest minute, Arnie has been riding for approximately 3 
hours and 45 minutes. Because Arnie started riding at noon, the time at which 
he and Barbara are 50 miles apart is approximately 3:45 pm. 



Answer: 9:42 am 



□ 



580 CHAPTER 8. QUADRATIC FUNCTIONS 



f> t* t* Exercises ■** •** ■** 

In Exercises 1-8, solve the given equation by factoring the trinomial using the ac-method, then applying 

the zero product property. Secondly, craft a second solution using the quadratic formula. Compare 
your answers. 

1. x 2 - 3x - 28 = 5. x 2 - 2x - 48 = 

2. x 2 -Ax- 12 = 6. x 2 + 9a; + 8 = 

3. a; 2 -8a; + 15 = 7. a; 2 + x - 30 = 

4. a; 2 - 6a; + 8 = 8. a; 2 - 17a: + 72 = 



In Exercises 9-16, use the quadratic formula to solve the given equation. Your final answers must be 
reduced to lowest terms and all radical expressions must be in simple radical form. 

9. a; 2 - 7x - 5 = 13. a; 2 - 7a; - 4 = 

10. 3a; 2 - 3a; - 4 = 14. a; 2 - 5a; + 1 = 

11. 2a; 2 +a;-4 = 15. 4a; 2 - x - 2 = 

12. 2a; 2 + 7a;- 3 = 16. 5a; 2 + x - 2 = 



In Exercises 17-24, use the quadratic formula to solve the given equation. Your final answers must be 
reduced to lowest terms and all radical expressions must be in simple radical form. 

17. a; 2 - x- 11 = 21. a; 2 - 3a; - 9 = 

18. a; 2 - 11a; +19 = 22. a; 2 - 5a; - 5 = 

19. a; 2 - 9a; + 9 = 23. a; 2 - 7a; - 19 = 

20. a; 2 + 5a; - 5 = 24. a; 2 + 13a; + 4 = 



In Exercises 25-32, use the quadratic formula to solve the given equation. Your final answers must be 
reduced to lowest terms and all radical expressions must be in simple radical form. 

25. 12a; 2 + 10a; - 1 = 29. 2a; 2 - 12a; + 3 = 

26. 7a; 2 + 6a; - 3 = 30. 2a; 2 - 6a; - 13 = 

27. 7a; 2 - 10a; + 1 = 31. 13a; 2 - 2a; - 2 = 

28. 7a; 2 + 4a; - 1 = 32. 9a; 2 - 2a; - 3 = 



•A. THE QUADRATIC FORMULA 



581 



33. An object is launched vertically and its 
height y (in feet) above ground level is 
given by the equation y = 240 + 160£ — 
16i 2 , where t is the time (in seconds) that 
has passed since its launch. How much 
time must pass after the launch before 
the object returns to ground level? Af- 
ter placing the answer in simple form and 
reducing, use your calculator to round the 
answer to the nearest tenth of a second. 

34. An object is launched vertically and its 
height y (in feet) above ground level is 
given by the equation y = 192 + 288£ — 
IQt 2 , where t is the time (in seconds) that 
has passed since its launch. How much 
time must pass after the launch before 
the object returns to ground level? Af- 
ter placing the answer in simple form and 
reducing, use your calculator to round the 
answer to the nearest tenth of a second. 

35. A manufacturer's revenue R accrued from 
selling x widgets is given by the equation 
R = 6000a; — 5a; 2 . The manufacturer's 
costs for building x widgets is given by the 
equation C = 500000 + 5.25a;. The break 
even point for the manufacturer is defined 
as the number of widgets built and sold so 
the the manufacturer's revenue and costs 
are identical. Find the number of widgets 
required to be built and sold so that the 
manufacturer "breaks even." Round your 
answers to the nearest widget. 

36. A manufacturer's revenue R accrued from 
selling x widgets is given by the equation 



R = 4500a; — 15.25a; 2 . How many widgets 
must be sold so that the manufacturer's 
revenue is $125,000? Round your answers 
to the nearest widget. 

37. Mike gets on his bike at noon and begins 
to ride due north at a constant rate of 6 
miles per hour. At 2:00 pm, Todd gets on 
his bike at the same starting point and be- 
gins to ride due east at a constant rate of 
8 miles per hour. At what time of the day 
will they be 60 miles apart (as the crow 
flies)? Don't worry about simple form, 
just report the time of day, correct to the 
nearest minute. 

38. Mikaela gets on her bike at noon and be- 
gins to ride due north at a constant rate of 
4 miles per hour. At 1:00 pm, Rosemarie 
gets on her bike at the same starting point 
and begins to ride due east at a constant 
rate of 6 miles per hour. At what time 
of the day will they be 20 miles apart (as 
the crow flies)? Don't worry about simple 
form, just report the time of day, correct 
to the nearest minute. 

39. The area of a rectangular field is 76 square 
feet. The length of the field is 7 feet longer 
than its width. Find the dimensions of the 
field, correct to the nearest tenth of a foot. 

40. The area of a rectangular field is 50 square 
feet. The length of the field is 8 feet longer 
than its width. Find the dimensions of the 
field, correct to the nearest tenth of a foot. 



582 CHAPTER 8. QUADRATIC FUNCTIONS 

41. Mean concentrations of carbon dioxide over Mauna Loa, Hawaii, are gathered by the Earth 
System Research Laboratory (ESRL) in conjunction with the National Oceacnic and Atmosphere 
Administration (NOAA). Mean annual concentrations in parts per million for the years 1962, 
1982, and 2002 are shown in the following table. 



Year 



Concentration (ppm) 
A quadratic model is fitted to this data, yielding 



1962 1982 2002 



318 341 373 



C = 0.01125r + 0.925£ + 318, 

where t is the number of years since 1962 and C is the mean annual concentration (in parts per 
million) of carbon dioxide over Manuna Loa. Use the model to find the year when the mean 
concentration of carbon dioxide was 330 parts per million. Round your answer to the nearest 
year. 

42. The U.S. Census Bureau provides historical data on the number of Americans over the age of 85. 



Year 



Population over 85 (millions) 
A quadratic model is fitted to this data, yielding 



1970 1990 2010 



1.4 3.0 5.7 



P = 0.01375i 2 + 0.0525* + 1.4, 



where t is the number of years since 1970 and P is number of Americans (in millions) over the 
age of 85. Use the model to find the year when the number of Americans over the age of 85 was 
2,200,000. Round your answer to the nearest year. 



j*- j*- j*- Answers •*$ **s ■*$ 



1.-4,7 ,l±v^$3 

15. 



17. 



3. 3, 5 

5. -6, 8 

7. -6, 5 19 



1±3V5 
2 

9 ±3^ 
2 

7± ^9 3±3V5 



9 

2 



2 



-1±a/33 7 ±5^5 

U * —^~ 23 - —2— 

13. l±M 25 . - 5± ^ 

2 12 



8.4. THE QUADRATIC FORMULA 583 

5_±_3_y/2 35. 90 widgets, 1109 widgets 

7 

6±V30 37 - 7:12 pm 

2 
^- 39. 5.9 by 12.9 feet 

31. — 

13 

41. 1973 
33. 11.3 seconds 



27. 



29. 



Index 



aomethod, 414 

graphing calculator, 462 

revised, 417 
^-intercept, 234 
^-intercept and zero of a function, 

325 
y-intercept, 201, 234 

absolue value, 20 

absolute value, 4, 16 

adding fractions, 36 

adding integers, 5 

additive identity property, 2 

algebraic expression, 21 

algebraic expressions, 60 

applications, 121 

approximating square roots, 542 

area 

rectangle, 471 
area of parallelogram, 119 
area of rectangle, 118 
area of trapezoid, 118 
area of triangle, 116 
ascending powers, 309 
associative property of addition, 7 
associative property of multiplica- 
tion, 9, 60 

binomial, 309 
brackets, 16 

calculator submission guidelines, 175, 
408, 419, 432, 520, 566 



canceling, 100 
cancellation, 34 

Cartesian Coordinate System, 152 
circumference of circle, 119 
clearing decimals, 99, 106 
clearing fractions, 99, 101, 113, 516 
clearing negative exponents, 485 
coefficient, 308 
combining like terms, 64 
commutative property of addition, 

6 
commutative property of multipli- 
cation, 9 
complementary angles, 282 
completing the square, 558 

procedure, 562 

solving equations, 563 
consecutive even integers, 124, 470 
consecutive integers, 122 
constant of proportionality, 525, 527 
constant rate, 184 
cost, 339 

decimals, 48 

addition, 49 

division, 51 

multiplication, 51 

rounding, 55 

subtraction, 49 
degree, 308, 311 
demand, 363 
dependent variable, 183 



585 



586 



INDEX 



descending powers, 309 
difference of squares, 373, 444 
distributing a negative sign, 63 
distributive property, 61 
distributive property for division, 511 
dividing fractions, 35 
divisor, 386 
domain, 294 

elimination method, 273 
equation, 74 

equivalent, 75 
solution, 74 
equations 

clearing decimals, 99, 106 
clearing fractions, 99, 101, 113 
different from expressions, 74 
equivalent equations, 75 
isolate terms containing x, 91 
meaning of "solve for x" , 90, 

111 
operations producing equivalent 

equations, 77, 79, 80 
simplify using order of opera- 
tions, 92 
equations in two variables, 156 

solution, 157 
equations with more than one vari- 
able, 111 
equivalent equations, 75 

adding same amount to both 

sides, 77 
dividing both sides by same amount, 

80 
multiplying both sides by same 

amount, 79 
subtracting same amount to both 
sides, 77 
equivalent inequalities, 139 
evaluating algebraic expressions, 21 
exponents 

clearing negative exponents, 485 
definition, 10, 343 
dividing like bases, 481 
multiplying like bases, 480 
negative, 478 



raising a power to a power, 481 

raising to —1, 478 

raising to a negative integer, 

482 
raising to negative integer, 479 

factor, 386 
factoring 

ax 2 + bx + c, a = 1, 413 

ax 2 +bx + c,a^l, 427 

applications, 467 

by grouping, 394 

checking the solution, 457 

definition, 390 

difference of squares, 444 

factor out GCF, 390 

first rule of factor, 430 

first rule of factoring, 390, 443, 
446 

perfect square trinomials, 442, 
561 

special forms, 440 

strategy, 456 
factoring completely, 446, 456 
FOIL method, 370, 372 
formulae, 111 
fractions 

addition, 36 

canceling, 100 

division, 35 

multiplication, 33 
free fall, 526 
function 

definition, 297 

rule of three, 301 
function notation, 301 
functions, 294 

y and f(x) are equivalent, 303 

adding, 334 

sketching the graph, 304 

subtracting, 337 

using different letters for dif- 
ferent functions, 303 

zero, 325 

geometric formula 



INDEX 



587 



sum of angle of a triangle, 126 
geometric formulae, 115 

area of a circle, 362 

area of parallelogram, 119 

area of rectangle, 118 

area of trapezoid, 118 

area of triangle, 116 

circumference of circle, 119 

perimeter of rectangle, 115 
graph of an equation, 157 
graphing 

scaling each axis, 183 
graphing an equation 

guidelines, 160 
graphing calculator, 171 

ac-method, 462 

adjusting the viewing window, 
176 

adjusting WINDOW parame- 
ters, 256 

changing system solutions to frac- 
tions, 267 

check solutions of rational equa- 
tion, 518 

checking fractional solutions, 105 

checking solutions, 78, 92 

checking solutions of equations, 
89 

checking solutions of systems, 
264 

checking solutions to nonlinear 
equation, 403 

checking solutions with radicals, 
560, 574 

drawing parabola, 314 

drawing polynomial of degree 
four, 316 

evaluating expressions, 23 

evaluating functions, 302 

evaluating polynomial function, 
323 

finding zero of a function, 326 

fractions, 39 

intersect utility, 253, 406, 418, 
432, 449, 540 

mode button, 171 



negation versus subtraction, 11 

plotting equation, 172 

reproducing results on home- 
work, 174 

rounding decimals, 55 

scientific notation, 497 

solving x 2 = a, 540 

solving nonlinear equations, 405 

solving rational equations, 519 

solving systems, 253 

STAT PLOT, 172 

submissioin guidelines, 175 

table feature, 160 

TRACE button, 178 

WINDOW menu, 176 

zero utility, 326, 407, 421, 435, 
565 
graphing equations by hand, 152 
graphing equations in two variables, 

157 
greater than, 136 
greater than or equal to, 136 
greatest common divisor, 31, 386, 

387 
greatest common factor, 386, 387 

two monomials, 389 
grouping symbols, 16, 19 

nested, 20 
guidelines for graphing an equation, 
160 

Hooke's Law, 112, 526 
horizontal line, 237 
hypotentuse, 549 

independent variable, 183 
inequalities, 135 

add or subtract same amount, 

140 
clearing decimals, 145 
clearing fractions, 144 
multiply or divide by positive 

number, 140 
multiplying or dividing by neg- 
ative number, 142 
reversing inequality sign, 141 



588 



INDEX 



integers, 2, 3 

addition, 5 

multiplication, 8 

subtraction, 7 
intercept, 201 
intercepts, 234 
interest, 286 
intersect utility, 406, 418, 432, 449, 

540 
interval notation, 138 

eyes "left to right" , 139 

summary table, 146 
inverse of addition, 76 
inverse of division, 80 
inverse of multiplication, 80 
inverse of subtraction, 76 
irrational numbers, 135 

kinetic energy, 114 

laws of exponents, 343, 480, 484 
dividing like bases, 346 
multiplying like bases, 344 
raising a power to a power, 348 
raising a product to a power, 

349 
raising a quotient to a power, 
351 

least common denominator, 36, 508 
procedure for finding, 510 

legs, 549 

less than, 136 

less than or equal to, 136 

light intensity, 528 

light-year, 500 

like terms, 64, 65 

linear equations 

strategy for solving, 401 

linear equations versus nonlinear equa- 
tions, 400 

list of squares, 543 

lowest terms, 31 

mapping diagram, 296 

notation, 299 
mass on a spring, 526 



measuring change in variable, 185 

mixture, 288 

monomial, 308 

multiplicative identity property, 10 

multiplying fractions, 33 

multiplying integers, 8 

multiplying radicals, 547 

natural numbers, 2 
negating a sum, 336 
negative exponents, 478 
nested grouping symbols, 20 
Newton's Law of Gravitation, 114 
Newton's law of gravitation, 500 
nonlinear equaitions 

graphing calculator solution, 404 
nonlinear equations, 399 

graphing calculator solution, 417, 
431, 447 

solving by factoring, 404, 417, 
431, 447 

strategy for solving, 402 
nonlinear equations versus linear equa- 
tions, 400 
number line, 135 
numbers 

integers, 3 

irrational, 135 

natural, 2 

rational, 31, 135 

real, 135 

whole, 2 

Ohm's Law, 113 

order of operations, 16, 37, 52, 66 

rules, 16, 37, 52, 66 
ordered pair, 152 
ordered pairs 

plotting, 154 
ordering real numbers, 136 
origin, 153 

parabola, 313 

opens downward, 314 
opens upward, 314 
vertex, 314 



INDEX 



589 



parentheses, 16 

perfect square trinomial, 442 

perimeter, 283 

rectangle, 471 
perimeter of rectangle, 115 
perpendicular, 235 
perpendicular lines, 219 
plotting ordered pairs, 154 
point-slope form, 213, 214 

applications, 222 

determine equation of line through 
two points, 214 

parallel lines, 217 

perpendicular lines, 220 

to standard form, 232 
polynomial, 309 

ascending powers, 309 

binomial, 309 

coefficient, 308 

degree, 308, 311 

degree four, 316 

graph, 313 

monomial, 308 

quadratic, 313 

reading the graph, 321 

term, 308 

trinomial, 309 
polynomial function, 312 

evaluating, 321 
polynomials, 308 

adding, 334 

adding and subtracting, 334 

applications, 321 

applications of addition and sub- 
traction, 338 

applications of multiplication, 
362 

descending powers, 309 

distribution shortcut, 360 

division of polynomial by a mono- 
mial, 511 

monomial products, 356 

multiplication, 356 

multiplying a monomial and poly- 
nomial, 357 

multiplying polynomials, 359 



negation, 335 

subtracting, 337 
prime factorization, 387, 510 

finding greatest common divi- 
sor, 388 
prime number, 32 
projectile motion, 323, 328, 468, 576 
property 

additive identity, 2 

associative, 7, 9, 60 

commutative, 6, 9 

distributive, 61 

multiplicative identity, 10 
proportional, 525 
proportionality constant, 525, 527 
Pythagorean Theorem, 549, 551, 577 

applications, 553 

proof, 550 

quadrants, 153 
quadratic equation, 571 
quadratic formula, 571, 572 
quadratic polynomial, 313 

radical expressions 

simplifying, 547 
radical notation, 536 
range, 294 
rate, 183 

constant, 184 
rational equation 

clearing fractions, 516 
rational equations 

applications, 521 

checking solution with graph- 
ing calculator, 518 

solving, 516 

solving with graphing calcula- 
tor, 519 
rational expression, 505 

adding, 507 

dividing, 506 

least common denominator, 508 

multiplying, 505 

subtracting, 508 
rational numbers, 31, 135 



590 



INDEX 



real numbers, 135 

ordering, 136 
reciprocal, 35, 521 
reducing fractions, 31 
relation, 294 

domain, 294 

range, 294 
Rene Descartes, 152 
reproducing calculator results on home- 
work, 174 
requirements for word problem so- 
lutions, 121 
revenue, 339, 363 
right angle, 549 
right triangle 

hypotenuse, 549 

legs, 549 
rounding, 55 
rule of three, 301 

scaling each axis, 183 
scientific notation, 492 
form definition, 495 
graphing calculator, 497 
how to report your answer, 499 
multiply decimal numbers by 
negative power of ten, 494 
multiply decimal numbers by 
nonnegative power of ten, 
493 
negative powers of ten, 492 
nonnegative powers of ten, 492 
placing a number in scientific 
notation, 495 
set-builder notation, 136 
summary table, 146 
simple radical form, 548 

factor out perfect square, 548 
simplifying radical expressions, 547 
slope, 183 

determining slope from graph 

of the line, 193 
downhill, 190 
drawing a line with given slope, 

194 
horizontal line, 191 



independent of direction of sub- 
traction, 189 

independent of selected points, 
187 

parallel lines, 217 

perpendicular lines, 220 

steepness of line, 189 

summary table, 195 

uphill, 189 

vertical line, 191 
slope as rate, 185 

slope of line through two points, 188 
slope-intercept form, 201, 202 

applications, 206 

determine equation from graph, 
205 

sketching line from given equa- 
tion, 202 

to standard form, 230 
solution of an equation, 74 
solution of equation in two variables, 

157 
solution of linear system, 246 
solve for x, 111 
solving equations 

completing the square, 563 

multiple steps, 87 

one step, 74 
solving rational equations, 516 
special product 

FOIL method, 372 
special products, 370 

difference of squares, 373 

FOIL method, 370 

squaring a binomial, 375, 378 
speed of light, 501 
square root 

negative, 537 

nonnegative, 537 
square roots 

approximations, 542 

definition, 537 

list of squares, 543 

multiplication property, 547 

simple radical form, 548 



INDEX 



591 



squaring a binomial, 375, 378, 440, 
561 

standard form, 228 
definition, 230 
identifying intercepts, 234 
to slope-intercept form, 228 

substitution method, 262, 263 

subtracting integers, 7 

sum of angles of a triangle, 126 

systems 

applications, 282 
checking solution with graph- 
ing calculator, 264 
infinite number of solutions, 252 
infinite solutions, 269, 277 
interpretation tip, 269 
no solutions, 250, 268, 276 
number of solutions, 253 
parallel lines, 250 
same lines, 252 
solution, 246 

solving by elimination, 273 
solving by graphing, 246 
solving by substitution, 262, 263 
solving using graphing calcula- 
tor, 253 



inverse, 527 
vertex of parabola, 314 
vertical line, 237 

whole numbers, 2 
word problem 

investment, 127 
word problems, 121 

requirements, 121 
wrap and unwrap, 75, 87 
writing mathematics, 17, 82 

zero of a function, 325 
zero product property, 399 
zero utility, 407, 421, 435, 565 
zeros and ^-intercepts, 406 



table feature of graphing calculator, 
160 

term, 308 

tips for evaluating algebraic expres- 
sions, 21 

tips for using function notation, 302 

trinomial, 309 

unit price, 363 
unlike terms, 64 



variable, 21 

change, 184 
dependent, 183 
independent, 183 
maintaining case, 111 

variable part, 64 

variation, 525 
direct, 525